Questions 245 and 246
245. (B) Ohmic materials will have a constant ratio of current to potential difference. This shows up on a graph as a straight line.
246. (D) In parallel, both bulb and resistor receive the same 10 volts.
247. (C) P = IΔV. Calculate power for both and add.
248. (B) R = ΔV/I. The resistance is 10 Ω for the resistor and 14.3 Ω for the bulb. Adding them in parallel gives us 5.9 Ω.
249. (B) This is a tricky one because the resistance of the bulb is non-ohmic, and its resistance changes with voltage and current. We know that in series the current passing through the resistor and bulb must be the same. Looking back at our graph, we see that this occurs at a current of 0.50 A. Checking to make sure that the voltages work out, we see that the potential across each will be 5 V, which works out perfectly because in series, the voltage drops of the resistor and the bulb must add up to the voltage supplied by the battery.
250. (A and D) Applying Kirchhoff’s loop rule to the top loop, we get answer choice A. Applying Kirchhoff’s junction rule to the right junction, we get answer choice D.
251. (A) The emfs of both batteries point in the same direction for the outer loop, for a combined potential difference of 15 V. The emfs point in the opposite direction for the line that contains R2, for a combined potential difference of 3 V.
252. (D) Originally only bulb A is lit, and it experiences all the emf of the battery. When the switch is opened, the emf of the battery is split evenly between the bulbs. therefore, the power dissipation by bulb A is one-quarter of the original.
Questions 253 and 254
253. (B) The three resistors in the bottom right corner are in parallel. Since they all have the same resistance and the same electric potential across them, they must also have the same current. Ammeter #3 is the sum of the currents in the top two resistors and will be twice as large as ammeter #4.
254. (B) The three resistors in parallel add up to a resistance of ⅓R. Adding these in series with the resistor in the main line, we get 4/3R.
255. (D) The resistor on the right has twice the radius and a quarter of the resistance of the one on the left. Therefore, the current through the right resistor will be four times larger.
256. (C and D) Measuring the potential difference when the switch is open gives us the battery’s emf. All we need is either the current or the potential difference, when the switch is closed, to find the internal resistance of the battery. With either, we can calculate the potential difference across the internal resistance and the current passing through it. Then we can find the value of the internal resistance.
257. (B) The reading in ammeter 1 never changes because the potential difference and resistance in that line does not change. The capacitor behaves like a wire when uncharged and like an open switch when fully charged. This behavior changes the current through ammeter 2 as time passes.
258. (C and D) Four capacitors connected in series result in a capacitance one-quarter the size of the individual capacitors. To get a single capacitor with one-quarter the capacitance, we need to get a factor of four in the denominator.
Questions 259 and 260
259. (A and D) The batteries are connected in series; therefore, their electric potentials add up. The difference between the voltages in the left graph is 1.5 V. Thus, the batteries must be 1.5 V each. The right graph shows power proportional to the voltage squared because the graph is linear.
260. (D) The slope of the right graph will equal the reciprocal of the resistance.
Questions 261 and 262
261. (B) The energy supplied to the gas is related to the power of the circuit, time, and the internal energy of the gas:
262. (B) The two resistors in parallel will have half the total resistance which will double the current and the power output of the circuit. This will double the slope of the graph.
AP-Style Free-Response Questions
263. (a) i. Minimum equipment: wires, voltmeter, and ammeter
ii. There are several ways to draw this. The key is that the ammeter is in series with the battery and the graphite. The voltmeter must be in parallel with the graphite to measure the correct potential difference. The graphite should be drawn as a resistor and the entire schematic needs to be labeled.
1. Connect the graphite, ammeter, and battery in series.
2. Connect the voltmeter in parallel with the graphite.
3. Measure the current and voltage.
4. Repeat for several lengths of graphite.
iv. Since the current and voltage are measured for the graphite specifically, the details of the emf source are irrelevant.
(b) i. Trials: 1, 6, 9, 11, 12, and 13 are the best because they give the widest quantity and spread of data for a constant length segment of Play-Doh with differing diameters. The resistance is missing and needs to be calculated. (See the table.)
ii. The graph should be a curve, and the axes must be labeled.
iii. It appears that resistance is inversely proportional to the diameter of the conductive path. But it is hard to tell. It could also be that resistance is inversely proportional to the diameter squared. Either answer would be appropriate based on the question and the information on display in the graph.
iv. To prove which relationship we actually have, we need to graph R − vs− 1/d and R − vs − 1/d2. Whichever graph shows a straight line indicates the correct relationship.
v. The changing water content of Play-Doh could change the resistivity of the material and the resistance-diameter relationship, which damages the validity of the lab. It is a variable that is not held constant. That is a problem.
264. (a) Agree. Writing Kirchhoff’s loop rule equation for the upper loop, we get
Applying Kirchhoff’s loop rule to the outer loop, we get
Combining the equations, we have
(b) P1 > P4 > P2 = P3. Power is P = I2R. All the resistors are the same. Therefore, the ranking is based on the current. Resistor 1 receives the most current as all the current must pass through it, and we just proved in answer (A) that resistor 4 receives twice the current of resistors 2 and 3.
(c) The current 1 in terms of current 3:
The power of resistor 1 in terms of resistor 4:
(d) Agree that the power will go down for resistor 1: When the switch is opened, there is only one path left for the current to pass through. This means the total resistance of the circuit increases. The potential difference across resistor 1 will decrease, which will bring its power dissipation down as well.
Disagree that resistor 2 and 3 are unaffected: The original current passing through resistors 2 and 3 was
The new current through resistors 2 and 3 is
The new current is larger than the old. Power dissipation goes up.
(e) i. Immediately after the switch is closed, the capacitor acts like a short circuit wire. The current through resistor 1: .
The potential difference across the capacitor: ΔV = 0.
ii. After a long period of time, the capacitor becomes fully charged and acts like an open switch in the circuit. The current through resistor 1: .
The potential difference across the capacitor will be equal to that of resistors 2 and 3 combined, because the capacitor is in parallel with them: .
iii. Potential energy:
265. (a) VT = ɛ − Ir
(b) i. It is important to use data from the best fit line of the graph. All experimental data have errors. Therefore, the data in the table have experimental errors. The best fit line is the best average of the data.
The easiest way to do this is to realize that the negative of the slope of the line is the internal resistance of the battery. Taking points from the best fit line, we get an internal resistance of 1.5 Ω. With this, we can pull a point off the graph and use the equation for part (a) to solve for emf = 12 V. We can also find the internal resistance by using data from the graph to set up two simultaneous equations with unknowns of ε and r.
ii. The y-intercept of the graph is the emf of the battery.
iii. The x-intercept is the maximum current that can be produced by the battery.
(c) i. Internal resistance of the battery will not influence the energy stored in the capacitor. When the capacitor is fully charged, there will be no current flowing from the battery. At this point, the graph shows that the terminal voltage of the battery is equal to the emf. Thus, the internal resistance has no effect on the voltage across the capacitor.
ii. Capacitors in parallel add up; therefore, the combined capacitance is 2C. The energy stored will be
iii. We need twice the capacitance of the original capacitor. We can accomplish this by doubling the plate area or cutting the distance between the plates in half: .
Chapter 5: Magnetism
266. Similarities: There are two types of charge (positive and negative) and two poles (north and south). Opposites attract, and likes repel. Both charges and poles produce fields that influence other charges and poles. Both fields decrease with distance.
Differences: Charges can exist individually, but poles always come in pairs (north and south). Charges exert forces on all other charges, while poles only influence moving charges. Electric forces are always parallel to the field lines. Magnetic forces are perpendicular to the field lines.
267. It will produce two smaller, weaker magnets.
268. Magnetic fields are produced by moving charges. Magnetic fields always come in “loops,” without beginning or end. This always produces a magnetic pair of poles.
269. The north end of a compass is attracted to the south magnetic pole of the Earth. The south magnetic pole of the Earth is in the geographic north (up where Santa lives). You can also say that the compass aligns with the Earth’s magnetic field with the north end of the compass pointing in the direction of the magnetic field. The Earth’s magnetic field exits the Earth in Antarctica and reenters near Canada and Russia.
273. Both are attracted to the magnet, but not because the bar is a magnet. The positive spheres polarize the magnet and are then attracted to it. The same thing would happen if it were a block of wood or metal instead of a magnet.
275. See figure. The arrow should be longest at A and shortest at B.
277. If you grasp a current-carrying wire with your right hand and the thumb pointing in the direction of the current, your fingers will curl around the wire in the direction of the magnetic field vectors.
278. Magnetic field is inversely proportional to the perpendicular distance from the wire.
279. There are actually several different ways to use this right-hand rule. Here is the one that most of the current textbooks seem to be using. Hold your hand flat with your thumb at a right angle to your fingers.
• Your fingers should point in the direction of the magnetic field. There are usually lots of magnetic field lines, and you have lots of fingers.
• Your thumb will point in the direction of the current in the wire.
• Perpendicular to the palm of your hand is the direction of the force on the current-carrying wire. A nice way to think of this is that the direction of the force on the wire is the same direction in which you would push on something with the palm of your hand.
281. (A) The bottom wire produces a magnetic field out of the page, in the region of the top wire, which causes a force directed downward toward the lower wire.
(B) According to Newton’s third law, the force is equal in strength but opposite in direction.
(C) The magnetic field is zero. The bottom wire produces an out-of-the-page magnetic field between the wires, while the top wire produces an into-the-page magnetic field. At the midpoint between the wires, the fields are equal in strength but opposite in direction. They cancel out.
282. (A) The bottom wire produces a magnetic field into the page, in the region of the top wire, which causes a force directed upward toward the top of the page and away from the lower wire.
(B) According to Newton’s third law, the force is equal in strength but opposite in direction.
(C) The magnetic field from the top wire at the midpoint is into the page:
The magnetic field from the bottom wire at the midpoint is also into the page: .
Because the fields are in the same direction, the net magnetic field is also into of the page:
(D) The magnetic force on the top current-carrying wire is due to the magnetic field of the lower wire: Therefore, the force per unit length of wire is
283. (A) Upward. See figure.
(B) The force will be doubled: FM = Il(sinθ)B.
(C) The force is zero because the current in the bottom wire of the circuit and the magnetic field are parallel (sinθ = 0).
284. There are several different ways to perform this experiment. Here is one example:
(A) Equipment: A long thin strip of aluminum foil, batteries, switches, wires with alligator clips, and a magnet with north and south ends marked
1. Suspend the thin strip of aluminum foil vertically; connect the top end to the positive end of a battery and the bottom end to the negative end with a switch.
2. Place the north end of the magnet close to the strip so the magnetic field passes perpendicular to the strip, as shown in the figure.
3. Close the switch and observe the direction in which the aluminum strip is deflected. This indicates the direction of the force on the current in the strip. Be sure to open the switch to stop the current, as you have produced a short circuit, which will produce a great deal of heat and drain the battery.
4. Now create another short circuit with a long wire, switch, and battery.
5. Suspend the wire parallel to the strip of foil.
6. Turn on both switches and observe the deflection of the foil. (Remember to open the switches to stop the current, as you have produced two short circuits, which will produce a great deal of heat and drain the battery.)
7. Using this method, you can investigate the effects of the magnetic field produced by a current-carrying wire by comparing it to the magnet’s effect on the foil strip.
285. Gravitational fields point inward toward the mass, causing an attractive force inward on all other masses.
Electric fields radiate outward from positives and inward toward negatives. Positive charges experience a force in the direction of the field, while negative charges experience forces opposite to the field.
Magnetic fields are produced by moving charges according to the right-hand rule. Only moving charges experience magnetic forces. Moving positive charges experience a force perpendicular to the magnetic field, consistent with the right-hand rule. Moving negative charges receive a force in the opposite direction to what is predicted by the right-hand rule.
286. There are actually several different ways to use the right-hand rule. Here is the one that most of the current textbooks seem to be using. Hold your hand flat with your thumb at a right angle to your fingers.
• Your fingers should point in the direction of the magnetic field. There are usually lots of magnetic field lines, and you have lots of fingers.
• Your thumb will point in the direction of the velocity of the particle.
• Perpendicular to the palm of your hand is the direction of the force on a positive moving particle. (Negative moving particles will receive a force in the opposite direction.) A nice way to think of this is that the direction of the force on the particle is the same direction in which you would push on something with the palm of your hand.
287. The direction of the forces are shown in the figure.
(A) FM = qv(sinθ)B = evB = 3.2 x 10-13 N.
(B) The force is the same magnitude: 3.2 x 10-13 N.
288. (A) Out of the page.
(B) The force is zero because the velocity and field are parallel: FM = qv(sinθ)B = 0.
(C) The force is out of the page. Since part of the velocity is perpendicular and part is parallel to the field, the electron will travel in a helical path circling along the magnetic field lines toward the bottom of the page.
289. Note that the neutron travels along a straight path. All positive particles curve toward the top of the page. The proton, electron, and positron all have the same charge, velocity, and magnetic force acting on them. Therefore, the lightest particles will curve with a tighter radius. The positron and electron have the same radius of curvature. The alpha particle has twice the charge of the proton, which means it will receive twice the force. However, it is also four times the mass of the proton. Therefore, it will have a greater momentum per magnetic force and will curve less.
290. (A) The fields are in opposite directions at point P. Remember to convert your distance to meters.
(B) The force is zero because the proton is not moving!
(C) FM = qv(sinθ)B = evB = 1.7 x 10-18 N toward the top of the page.
291. (A) The magnetic field is downward at the origin because the magnetic field from the 2A wire dominates. By the right-hand rule, the force on the proton will be into the page.
(B) The fields are in opposite directions to the right of 8 cm and will cancel out where the fields have the same magnitude. This is satisfied at 12 cm:
292. The bottom right and top left wires will produce magnetic fields directed upward to the right. The other two wires produce magnetic fields directed downward and to the right. The net sum of all the four individual magnetic fields is horizontally to the right.
293. (A) The magnetic field is to the left. By the right-hand rule, the force is out of the page.
(B) Into the page, by Newton’s third law.
(C) The force is proportional to the velocity; therefore, the force is doubled. The direction is the same.
294. (A) Remember that the magnetic force is the centripetal force:
(B) Rearranging our equation, we get mv = rqB.
(C) Rearranging our equation, we get .
(D) Remember that the velocity of an object in circular motion is the circumference divided by the time period:
Note that the time period is not dependent on the velocity of the charged particle.
295. The field should be uniform and constant, filling all the space between the plates. Due to the polarity of the battery, the top plate is positive. Therefore, the electric field is directed downward.
297. The particle will accelerate downward toward the bottom plate in a parabolic path.
298. We need a magnetic force upward. Using the right-hand rule, the magnetic field must point into the page.
299. The electric and magnetic forces must cancel out: FE = FM
300. The electric force remains the same, but the upward magnetic force increases because it is proportional to the velocity. Therefore, the particle will curve upward toward the top plate. Eq < q(2v)B.
301. Because the charge cancels out of the equation, the negative particle will travel straight through without curving: Eq = qvB.
302. (A) The magnetic force on the negative charge is to the left. We need an electric field to the left.
(B) The electric force on the electron is to the left. We need a magnetic field out of the page.
(C) The magnetic force is out of the page. We need an electric field into the page.
303. Use conservation of energy:
304. Remember that the electric and magnetic forces must cancel out. FE = FM
305. Remember that the magnetic force is the centripetal force.
306. Diamagnetic materials have no unpaired electrons in their electron shells. These materials produce a weak, repulsive magnetic field when placed in an external magnetic field. When the external magnetic field is removed, the internal repulsive field disappears. Carbon, copper, and gold are diamagnetic.
Paramagnetic materials have at least one unpaired electron in their electron shells. These materials produce a weak, attractive magnetic field when placed in an external magnetic field. When the external magnetic field is removed, the internal attractive magnetic field disappears. Sodium, oxygen, and aluminum are paramagnetic.
Ferromagnetic materials have multiple unpaired electrons in their electron shells that easily produce a strong net magnetic field in the material. They have naturally forming magnetic domains inside the material. These domains can be aligned to produce a permanent magnet. Iron, cobalt, and nickel are ferromagnetic.
307. Only ferromagnetic materials have magnetic domains. Magnetic domains are areas where groups of atomic magnetic fields are aligned in the same direction.
308. When exposed to an external magnetic field, ferromagnetic domains align with the external field, producing a very strong magnetic field in the same direction as the external field. When the external magnetic field is removed, the domains can remain aligned, forming a permanent magnet.
309. A diamagnetic fluid will be weakly repelled by a strong magnet. A paramagnetic material will be weakly attracted to a strong magnet.
310. Magnet flux occurs when magnetic field lines pass through an area.
311. ΦB = B(cosθ)A. To change the magnetic flux, all we need to do is change one of the variables in the equation. We can increase or decrease the magnetic field strength, change the angle between the area and the magnetic field, or change the area that is exposed to the magnetic field.
312. (A) There are many answers to this one. All we need to do is change one of the variables in the magnetic flux equation. Here are two examples: move the loop to the left, out of the magnetic field, or rotate the loop about a diameter.
(B) There are many answers. Here are two: move the magnet away from the loop, or drop the magnet through the loop.
(C) Here are two examples: rotate the magnet about its long axis as shown in the figure, or move both the magnet and the loop in the same direction at the same time, keeping the same distance between them, so there is no relative motion between them.
313. (A) ΦB = B(cosθ)A. Change the area by collapsing the loop of wire so it has a smaller cross-sectional area. Change the angle between the field and the loop by rotating the loop about a diameter. Change the field strength.
(B) The loop could be rotated clockwise or counterclockwise; alternatively, just move the loop to the right in the field without exiting the field.
(C) By Lenz’s law, the induced magnetic field inside the loop will oppose the change in magnetic flux. Since the field into the page is getting stronger, the current will be counterclockwise to produce an opposing field out of the page.
314. (A) There is a change in magnetic flux only at locations 2 and 4.
(B) The flux is increasing into the page at location 2. Therefore, the current is counterclockwise to produce an out-of-the-page field to oppose the increase. At location 4, the flux is decreasing. Therefore, the current is clockwise to produce an into-the-page field to oppose the decrease in the flux.
315. (A) ɛ = Blv = Byv
(C) The magnetic flux is increasing out of the page. Therefore, the current will be clockwise to produce a magnetic field into the page to oppose the increase in flux. Thus, the current will be upward through the resistor.
316. A microphone has a flexible membrane with a coil of wire attached that vibrates when sound waves strike the membrane. The coil sits near a magnet. The vibration causes changes in the magnetic flux that produces oscillating currents in the coil, which are then transmitted to a speaker to produce sound.
An electric generator consists of loops of wire attached to a rotating shaft. Turning the shaft swings the coils of wire past magnets. This produces a change in flux by changing the angle between the magnetic field vectors and the coils. The current induced in the wires is then transmitted from the generator to power electrical devices.
AP-Style Multiple-Choice Questions
317. (C) The original velocity is toward the top of the page. The force on the proton is out of the page. Therefore, by the right-hand rule, the magnetic field is directed to the left.
318. (C) The original velocity of the electron is toward the bottom of the page. The force on the electron is to the right. Therefore, by the right-hand rule, the magnetic field is out of the page. Remember that the electron is negative and receives a force opposite to what positive receives!
Questions 319 and 320
320. (D) . The mass of the electron is smaller; therefore, the radius is smaller as well. The charge is opposite, so the electric force is in the opposite direction.
321. (A and D) Be careful! Make sure you are paying attention to which of these is a magnetic field and which is an electric field. Electric forces are along the axis of the field. Magnetic forces abide by the right-hand rule.
322. (B) Newton’s third law.
323. (D) It is not possible to separate a north pole from a south pole. All magnets are dipoles. When you break a magnet in half, you get two weaker magnets. If they stayed the same magnitude as the original, we would be violating conservation of energy.
324. (C) The dipoles in ferromagnetic materials align with the external magnetic field and amplify it. The dipoles in a diamagnetic material do not align with the external magnetic field.
325. (B) Use the right-hand rule for magnetic fields around current-carrying wires.
326. (C) By the right-hand rule, the magnetic force on the proton is to the left. Originally, the electric and magnetic forces were equal. Since the velocity has increased, the magnetic force is now larger than the electric force that is to the right.
Questions 327 and 328
327. (A) The magnetic field, due to the current in the wire near the electron, is into the page. By the right-hand rule, the negative electron will receive a force to the left due to the current. By Newton’s third law, the force on the wire will be equal and opposite to the right.
328. (A and C)
329. (D) The diaphragm vibrates back and forth along the axis of the magnet, changing the magnetic field strength through the coil area.
AP-Style Free-Response Questions
330. (a) The capacitor is connected to the battery so the left plate is at a higher potential than the right plate. This will accelerate positive ions to the right and hinder electrons from crossing between the plates.
ii. This must be a circular path, as shown in the figure.
(c) The radius goes up by a factor of . A reasonable answer range is shown in the figure above.
(d) The velocity of the doubly ionized nitrogen will be increased by a factor of passing through the capacitor. Inside the magnetic field, the radius of motion is decreased by a factor of .
331. (a) The force of gravity of the electron is much smaller than the electric and magnetic forces. This is due to the mass of an electron being many times smaller than its charge, and the universal gravitational constant being many times smaller than Coulomb’s law and magnetic constants.
(b) i. The electric force and magnetic force cancel out. In the free body diagram, the forces must be equal, and the magnetic force must be upward, as shown in the figure.
iii. The bottom plate. The magnetic force on the electron is upward. Therefore, the electric force must be downward to cancel out the magnetic force. This means the electric field is directed upward, and the lower plate must be a higher potential.
(c) i. The electric force cancels out the magnetic force, and the charge drops out of the equations.
ii. The magnetic force will be greater than the electric force. The path of the positron should be curving downward, as shown in the figure.
(d) i. The moving charge is experiencing a magnetic force from the current-carrying wire:
The potential difference to resistance ratio is simply the current in the wire:
ii. The current is to the clockwise around the circuit. The magnetic field from the top wire is out of the page in the vicinity of the electron. The magnetic force vector on the electron is directed upward toward the top of the page. The force vector arrow should be drawn upward.
Chapter 6: Waves and Optics
332. A wave is a disturbance in a medium that transports energy through that medium.
333. Differences: Longitudinal waves vibrate parallel to the direction of wave propagation, while transverse waves vibrate perpendicular to the direction of wave propagation. Transverse waves can be polarized, but longitudinal waves cannot. All electromagnetic (EM) waves are transverse.
Similarities: Both waves transport energy and have a frequency, time period, and amplitude. Both exhibit interference, diffraction, refraction, reflection, and Doppler effect.
334. Here are several different representations of longitudinal waves:
Here is a representation of a transverse wave and an EM wave.
335. Differences: The biggest difference is that mechanical waves require a physical medium to transport through. Electromagnetic waves do not require a physical medium as they create their own electric/magnetic field and are said to be self-propagating. Thus, EM waves travel through vacuums, but mechanical waves do not. Mechanical waves can vibrate as longitudinal waves or transverse waves, depending on how they are created and the physical medium they pass through. All EM waves are transverse, with both a transverse electric field wave and a transverse magnetic field wave perpendicular to each other. Electromagnetic waves all travel at the speed of light c in a vacuum and “slow down” when moving through mediums. Mechanical waves travel much slower, at speeds determined by the medium, generally traveling fastest in solids and slowest in gases.
Similarities: Both exhibit the classic wave properties of interference, diffraction, refraction, reflection, and Doppler effect.
336. Note the wave equation is shown as a cos function, but it can also be a sin function.
338. (A) This is a cosine wave with an amplitude of 4.0 T and wavelength of 4 m.
(B) E = 2sin(1.05t). This is a sine wave with an amplitude of 2 N/C and wavelength of 6 s.
339. When waves occupy the same space, the net amplitude is the sum of the amplitudes of the separate waves.
340. Constructive interference: When two waves overlap in phase (crest to crest), the net amplitude increases.
Destructive interference: When two waves overlap out of phase (crest to trough), the net amplitude decreases.
341. Here are six of the interference patterns seen as the two waves overlap and then pass by each other.
342. The Doppler effect occurs when the source of the wave and the observer of the wave have a relative velocity between them. When the two are moving apart, the EM wave will have a longer wavelength and lower frequency (red shift). When the two are moving toward each other, the EM wave will have a shorter wavelength and higher frequency (blue shift). Examples of technology that use the Doppler effect are Doppler weather radar and police radar guns. Astronomers use red and blue shift to measure how fast stars are moving away from or toward the earth.
343. Electromagnetic waves are a vibration in electric and magnetic fields. The changing electric field produces the magnetic field, and the changing magnetic field produces the electric field. Thus, an EM wave can propagate itself through a vacuum.
344. Electromagnetic waves can be polarized, and only transverse waves can be polarized.
345. The rope has a vibration axis. When the rope axis is aligned with the axis of the picket fence, the wave passes through unchanged. However, when the axes are perpendicular, the wave cannot pass through the fence.
346. (A) Nonpolarized light is vibrating along all the axes equally. Thinking of this in terms of x-y components, this means that 50 percent is vibrating along the x-axis, and 50 percent is vibrating along the y-axis. Therefore, a vertical polarizing filter will only let through the 50 percent of the light vibrating along the y-axis; it will block the light vibrating along the x-axis.
(B) The first filter blocks horizontal axis light. The second lets through horizontal axis light, but there is none left. Therefore, no light passes through the second filter.
(C) The first filter lets vertical axis light through. The second filter lets angled axis light through. The vertically polarized light has a component of light along the angled axis. Therefore, a component of the vertical light will pass through the second filter.
347. Radio, microwave, infrared, visible, ultraviolet, X-ray, gamma ray
348. Similarities: Both have the same method of propagation. Both are EM waves that travel at the speed of light.
Differences: The only difference is the wavelength/frequency and energy of the wave. Radio waves have longer wavelengths, lower frequencies, and less energy than X-rays.
349. c = 3.0 × 108 m/s in a vacuum.
350. 400 nm (violet light) to 700 nm (red light)
351. Using the equation, v = fλ, we get a frequency range of approximately 4.3 × 1014 Hz (red light) to 7.5 × 1014 Hz (violet light).
352. Wave fronts show where the wave crests and troughs are located. Rays indicate which direction the wave is traveling. Rays are always perpendicular to wave fronts.
353. Huygens’ principle states that each point on a wave front is the source of a wavelet that moves outward from the original wave front. The superposition of all the individual wavelets creates the new wave front.
354. (A) The portion of the wave that hits the boundary on the right side will be reflected or absorbed. The left half of the wave will pass by the boundary. The wavelets that are produced at the edge of the boundary will cause the wave to curve around the edge of the boundary.
(B) Sound waves have a long wavelength, and wavelets near the boundary will cause a pronounced diffraction effect. As the wavelength gets smaller than the boundary itself, as in visible light waves, the diffraction effect is less pronounced because the wavelets near the boundary are small.
(C) The point source model shows that when the opening is comparable in size to the wavelength, there will be a pronounced diffraction (bending of the wave front) around the corners of the opening.
(D) When the opening is much larger than the wavelength, the point source model shows that only the edges of the wave show any bending or diffraction. The wavelets in the center continue the propagation of the plane wave.
(E) The point source model shows that the two slits will produce two separate curved wave fronts passing through the openings. These two new wave fronts will interfere with each other, creating a pattern of constructive and destructive interference.
(F) The speed of light in the new medium is slower. Drawing the Huygens’ wavelets, we can see that all parts of the wave will slow down at the same time. This means the wave front will not change direction but will simply travel straight through the glass with a shorter wavelength.
(G) Entering the new medium, the wave slows down. Drawing the wavelets in the new medium with a shorter wavelength, we see that the wave front changes direction, and the wave ray bends toward the normal line to the surface.
(H) Drawing wavelets for the wave that reflects off the surface, we see that the wave maintains the same speed and wavelength. However, the reflected wave front turns and travel off in a new direction. The angle of the incoming wave front measured to the normal is equal to the angle of the outbound reflected wave front measured to the normal line.
355. The light and dark pattern of single-slit diffraction is described by this equation: dsinθ = mλ. Where d is the slit or opening width, λ is the wavelength of the light, θ is the angle from the slit to the minima (dark spot) on the screen, and m is the minima number. Note that m = 1 is the first minima to either side of the central maxima (bright spot), and m = 2 would be the second minima. Solving the equation for our situation, we get Therefore, as λ goes up or W goes down, θ increases.
(A) As X increases, there will be no change to the pattern because nothing in the equation changes.
(B) As Y increases, nothing in the equation changes. However, since the screen is farther away from the barrier, there is a greater distance for the pattern to spread out before it hits the screen. Therefore, the pattern will become wider or more spread out, as shown in the figure.
(C) When W increases, θ decreases. Therefore, the pattern becomes narrower and more tightly packed.
(D) Red light has a longer wavelength, which will increase θ, making the pattern spread out.
(E) Violet light has a shorter wavelength, which will decrease θ, making the pattern narrower.
356. This is a situation of single-slit diffraction: dsinθ = mλ. The deflection angle to the first-order minima is determined by the ratio of wavelength over opening width: The width of a door is roughly 1 m, which is similar in size to the wavelengths of sound. Therefore, the ratio and . This means the sound will diffract all the way around the corner and fill the entire room beyond. Light has a very small wavelength (400–700 nm). Therefore, the ratio . This means the light will diffract very little as it passes through the doorway because the doorway is so much larger than the wavelength of light. Light actually does diffract through the door—just not very much.
357. (A) The light and dark pattern of double-slit diffraction is described by this equation: dsinθ = mλ, where d is the distance between the two slits; λ is the wavelength of the light; θ is the angle from the slit to the maxima (bright spot) on the screen; and m is the maxima number. Note that m = 0 is the central maxima right down the center along the line of symmetry, and m = 1 would be the first maxima to either side of the central maxima. Solving the equation for our situation, we get Therefore, as λ goes up or Z goes down, θ increases.
(B) As W decreases, there will be no change to the pattern because nothing in the equation changes.
(C) As X decreases, there will be no change to the pattern because nothing in the equation changes.
(D) As Y decreases, nothing in the equation changes. However, since the screen is closer to the barrier, there is less distance for the pattern to spread out before it hits the screen. Therefore, the pattern will become closer together and more tightly packed.
(E) As Z decreases, θ increases. Therefore, the pattern becomes wider and more spread out.
(F) Red light has a longer wavelength, which will increase θ, making the pattern spread out.
(G) Violet light has a shorter wavelength, which will decrease θ, making the pattern narrower.
358. A bright spot will occur whenever there is constructive interference. This happens when the waves from the two openings reach the screen in phase (crest-crest or trough-trough). This will always occur in the center along the line of symmetry, because both waves travel the same distance to reach that location. This also occurs when the path length difference (ΔL) to a point on the screen is a multiple of the wavelength: ΔL = mλ. Therefore, there will be multiple locations of constructive interference corresponding to path length differences of ΔL = 1λ, 2λ, 3λ, and so on.
Dark spots will occur where there is destructive interference. This happens when the waves from the two openings reach the screen out of phase (crest-trough). This occurs when the path length difference (ΔL) to a point on the screen is a half-multiple of the wavelength: ΔL =mλ. Therefore, there will be multiple locations of destructive interference corresponding to path length differences of , and so on.
359. Similarities: Both patterns are described by the equation dsinθ = mλ, where d is the distance between the two adjacent slits; λ is the wavelength of the light; θ is the angle from the slit to the maxima (bright spot) on the screen; and m is the maxima number. Therefore, the pattern spacing is the same.
Differences: Double-slit maxima are spread out, while multislit diffraction grating maxima are very tight and point-like, as shown in the figure.
360. When light strikes a thin film, some light reflects off the top of the film. The rest travels through the film, and some of this reflects off the bottom of the film and back up through the film again. This transmitted light has traveled a distance equal to twice the thickness of the film. To produce constructive interference, the thickness will need to be ½λ so the path length difference is a whole multiple of the wavelength. This will cause the two waves to align in phase, creating constructive interference. The index of refraction of the coating will change the wavelength of the light in the coating. Thus, the thickness of the coating depends on the wavelength of the light in the coating, not in the air.
When light strikes a more optically dense medium, the reflected wave will be inverted. Inverted reflections are a phase change equivalent to the wave jumping ½λ. If the wave is inverted at both reflections, the effect cancels out. However, if only one inversion occurs, the thickness of the thin film will need to be only ¼λ thick because the interfering waves are already ½λ apart.
361. The law of reflection: The incoming angle = the reflected angle measured from the normal to the surface.
364. When light or any other wave encounters a boundary, the light can be reflected, refracted (transmitted), or absorbed. The more similar the substances, the more the light will be refracted through the new medium. The less similar the substances, the more the light will reflect off the surface. The total amount of light energy that is reflected, refracted, and absorbed will equal the original energy of the light striking the surface.
365. The velocity of the light and wavelength will change: The frequency will remain the same.
(C) The frequency in water is the same as in air:
369. The dots on the left are brighter both above and below the acrylic block. Conservation of energy tells us that less and less of the light is making it through to the right dots.
370. Ray #1 must be striking the acrylic/air interface at an angle of incidence larger than the critical angle. This produces total internal reflection along the parallel top and bottom surfaces, as shown by the dashed line in the figure.
371. Ray #2 must be striking the acrylic/air interface at an angle less than the critical angle. There is a partial reflection and refraction at each surface. This will produce a dot on each screen, as shown by a solid line in the figure from question 370.
Therefore, and angle equal to 49° or larger will not enter the air.
(D) No! Traveling from air into water, the light ray will slow down and turn toward the normal. There is no critical angle. Therefore, some of the light will always enter the water from the air. Even when the angle of incidence from the air is 90 degrees, some of the light will enter the water at an angle of 49 degrees.
373. (A) Equipment: protractor, pencil, ruler, sheet of white paper
1. Place the prism in the center of the sheet of paper.
2. Trace the outside of the prism with the pencil.
3. Direct the laser beam so it passes through the prism.
4. Trace the ray’s path with the ruler before it enters the prism and after it exits, being sure to mark the entrance and exit locations.
5. Remove the prism. Mark the normal line at the incoming and outgoing surfaces. Draw the light ray’s path through the prism.
6. Using the protractor, measure the angles of incidence and refraction for each incoming and outgoing ray.
7. Use Snell’s law to calculate the index of refraction of the glass prism.
8. Repeat for a wide variety of angles, and average the result.
374. (A) Use Snell’s law and the first set of data:
Repeating this for the rest of the data in the table, we get an average of 1.76.
(B) We can see in the figure that the trend line clearly shows the data are not straight and cannot be a direct relationship.
(C) Plotting the sine of the refracted angle as a function of the sine of the incident angle produces a straight line. This makes sense:
The slope of the line will equal the index of refraction of the sapphire. The best fit line gives a slope of 0.565. Thus, ns = 1.77.
375. (A) The incident ray partially reflects and refracts at the surface. This allows for multiple exit points from the prism.
(B) Note that all angles with the normal are smaller inside the prism than the rays in the air. This is because the light is traveling more slowly in the prism than in the air.
(C) Note that all the reflected angles in the preceding figure obey the law of reflection. The incoming and reflected rays are the same. This is shown with marks like the ones used in geometry class.
376. (A) Using Snell’s law:
(B) Using some geometry, we can deduce that the angle beta is 41 degrees. (Use the idea that the internal angles inside any triangle add up to 180 degrees.) Using Snell’s law, we see that the light ray does indeed exit the prism as shown in the figure:
377. (A) The minimum index of refraction so light does not escape at a critical angle of 45 degrees is shown by the following equation:
(B) When in air, the light strikes the bottom of the prism beyond the critical angle. When the prism is lowered into water, the difference in the index of refraction and speed of light between the two mediums is less than it was while in air. Therefore, the light does not speed up as much going into the water as it did going into air. This reduces the bending of the light due to refraction. The light is no longer hitting the bottom of the prism beyond the critical angle. Therefore, the light is able to escape out the bottom into the water but not the air.
379. Fiber optics consist of a thin transparent tube. Light shines in one end. Because the fiber is so thin, the angle of incidence inside the tube is always beyond the critical angle, and the light cannot escape until it reaches the other end of the fiber.
380. Both images can be seen. Real images are formed by light converging to a point and, thus, can be projected on a screen. Virtual images are formed by diverging light rays that appear to come from a specific point but have not. Virtual images cannot be projected on a screen.
381. The focal length of a spherical mirror is equal to half the radius of the mirror.
382. The focus, like the name implies, is the location where parallel light rays striking the mirror converge for concave mirrors and convex lenses. Parallel rays appear to diverge from the focus for convex mirrors and concave lenses.
383. Lenses A, C, and F are converging/convex lenses because they are thicker in the middle than at the edges. Lenses B, D, and E are diverging/concave lenses because they are thinner in the middle than at the edges. Mirror G is converging/concave. Mirror H is diverging/convex.
384. For converging/convex lenses:
1. A ray, parallel to the principal axis, refracts through the lens and passes through the focus on the opposite side of the lens.
2. A ray passing through the front focus and is refracted parallel to the principal axis on the other side of the lens.
3. A ray passes straight through the center of the lens, and its direction does not change.
For diverging/concave lenses:
1. A ray, parallel to the principal axis, refracts away from the principal axis as if it has come from the near-side focus.
2. A ray directed toward the far-side focus is refracted parallel to the principal axis.
3. A ray passes straight through the center of the lens, and its direction does not change.
385. See figure
386. See figure
387. See figure
388. See figure
389. The reflection from the mirrors will remain exactly the same, as the law of reflection is does not depend on the medium. The rays traveling through the lenses will refract less because there will be less of a velocity change when the light travels from water to lens.
390. The mirror’s focal length will not change as it depends only on the curvature of the mirror. However, lens optics depend on both the shape of the lens and refraction. The refraction of the lens will be less for oil/lens than for air/lens because the light will not change speed as much. Less refraction means the light rays will bend less going through the oil/lens, and the focal length will be greater than for the air/lens configuration.
391. (A) Note that the rays are closer to the normal line inside the lenses because the speed of light is slower inside the lens than in air.
(B) Increasing the thickness of the lens will increase the angle of incidence at which the incoming ray strikes the lens. This increases the refraction at the surface and the deflection of the inbound rays but decreases the focal length of the lens.
(D) Light rays converge to form real images. Light rays diverge from the location of a virtual image as if they came from that spot, even though they did not actually pass through that location.
393. (A-C) See figure.
(D) Light rays converge to form real images. Light rays diverge from the location of a virtual image as if they came from that spot, even though they did not actually pass through that location.
(E) Since half of the light rays will still be passing through the lens unimpeded, the image will still form in the same spot with the same properties as before. However, the image will not be as bright because half of the rays from the flower are now blocked.
395. The light rays will converge to the focus. Therefore, the focal length is 20 cm. The radius will be twice the focal length, or 40 cm.
396. Since the image is inverted, the lens must be a converging lens with a positive focal length:
The image is twice as large as the object and inverted. The object is 15 cm from the lens.
397. Since the image is upright and larger, the mirror must be concave, with a positive focal length creating a virtual image. The image is upright, so the magnification must be positive.
This makes sense because virtual images have a negative image distance.
The focal length is positive just as we expect for a concave mirror. The radius of curvature will be twice the focal length, 16.6 cm.
398. We need to be careful here! Diverging optical devices have a negative focal length, and we expect the image to have a negative image distance because it will be virtual. Since we have a diverging optical device, the magnification should be positive (upright) and less than one.
399. Eyes produce a real image on the retina, so the focal length should be positive.
400. Diverging optical devices have a negative focal length. The focal length of a mirror is half the radius of curvature; therefore, the focal length is −25 cm. Diverging optical devices only produce virtual images with negative image distances, so the distance to the image is −15 cm.
The magnification tells us that the image is upright and smaller than the object. This is what we would expect to get from a diverging optical device.
401. There are several ways to perform this lab. Here is one method.
(A) Equipment: mirror, meter stick, candle, screen
1. Place the mirror, candle, and screen on a table, as shown in the figure.
2. Move the screen until a crisp image forms.
3. Measure the object distance from the mirror to the candle and the image distance from the mirror to the screen.
4. Repeat this process for several data points.
5. Graphing on the x-axis and on the y-axis, we should get a graph with a slope of –1 and an intercept that equals (You can also use the mirror equation to calculate the focal length for each set of data and average, but the AP test usually asks you to graph a straight line graph to find what you are looking for.)
(D) No! A convex mirror will not produce a real image on the screen.
402. There are several ways to perform this lab. Here is one method.
(A) Equipment: lens, meter stick, candle, screen
1. Place the lens, candle, and screen on a table, as shown in the figure.
2. Move the screen until a crisp image forms.
3. Measure the object distance from the lens to the candle and the image distance from the lens to the screen.
4. Repeat this process for several data points.
5. Graphing on the x-axis and on the y-axis, we should get a graph with a slope of –1 and an intercept that equals
(Note: Another easy way to find the focal length is to take the lens outside on a sunny day. Produce a point of light with the rays from the sun. The distance from the lens to the point of light will be the focal length.)
(D) No! A concave lens will not produce a real image on the screen.
Questions 403 and 404
403. Graphing on the x-axis and on the y-axis, we should get a graph with a straight line with a slope of –1.
404. The intercept of the line equals
The intercept equals 0.1 (1/cm). Therefore, the focal length is 10 cm.
405. (A) The crystal ball is behaving like a convex/converging lens because it is inverting the image.
406. (A) The image formed on the retina is real. Therefore, the lens must be convex/converging.
(B) A concave/diverging lens is needed to spread the rays outward so that the eye can focus them on the retina.
(C) A converging/convex lens is needed to converge the light rays before they enter the eye. The lens in the eye converges the light the rest of the way to focus the image on the retina.
AP-Style Multiple-Choice Questions
407. (C) The law of reflection is not influenced by the water. Snell’s law of refraction depends on the indices of refraction of the two materials. The speed of light changes less going from water to lens then going from air to lens. This means there will be less refraction in water, making the focal length larger.
408. (B) Electromagnetic waves are transverse, with both the E- and B-fields oscillating perpendicular to the direction of motion and each other.
409. (A and B) . Increasing the wavelength of the laser (λ) and/or decreasing the hair width (d) will both increase the angle of the pattern.
410. (D) The point source model shows us that the larger the wavelength, the greater the bending of the wave around the corner. We can also see this in the equation for diffraction: . As the wavelength gets smaller, the diffraction bending angle also gets smaller. Visible light has a very small wavelength and only bends a tiny amount around corners.
411. 80. (C) The amplitude is 3 mV/m, and the time period is 0.5 ns. Using the wave equation, we get
412. (A) The image is upright and smaller. This means the image is virtual, the image distance is negative, and the mirror must be diverging/convex.
413. (A) The frequency remains the same. The wavelength of light is directly proportional to the speed of light in the substance. Light travels faster in water than through glass; therefore, λw > λg.
414. (D) The lens equation can be rearranged to produce a straight line:
y = mx + b
Thus, if we plot on the x-axis and on the y-axis, we should get a graph with a slope of –1 and an intercept of . The image distance is x2 – x1, and the object distance is x0 – x1.
415. (B) When the angle of the prism decreases, the right side of the prism becomes more vertical, and the angle of incidence with the normal becomes smaller. This creates less refraction and the distance (x) increases. Look at the extreme case when θ becomes zero. Then the angle of incidence is zero, and there is no refraction at all. The beam will pass straight through the “prism” because it has become flat like a window, and the distance (x) becomes infinite.
416. (B and D) Drawing the normal lines will make the paths more evident. The light ray should bend toward the normal when entering glass and bend away from the normal when entering air. Note that answer choice D shows light entering/exiting the prism along the normal and with total internal reflection inside the prism.
417. (A) Be careful of units! Convert the object distance from 30 cm to 300 mm.
This means the image is 3.1 mm in front of the retina, which is at a distance of 20 mm. We need a diverging lens to move the image back to the retina. Diverging lenses are concave.
AP-Style Free-Response Questions
418. (a) First, we need to find the object and image distance from the data in the table. Knowing that the lens is located at 50 cm and using the first set of data, we get the following:
s0 = 50 cm – 0 cm = 50 cm
s0 = 71 cm- 50 cm = 21 cm
The focal length can be calculated using the lens equation:
(We can confirm this data point by calculating the focal point with the other data points in the table. The last two columns will be used for part b).
(b) The lens equation can be rearranged to produce a straight line:
y = mx + b
Thus, if we plot on the x-axis and on the y-axis, we should get a graph with a slope of –1 and an intercept of From our graph, the intercept is 0.067 1/cm, which gives us a focal length of 15 cm (the same as part a).
(c) i. An upright image will be virtual. The object will need to between the lens and the focal point. Sketches will vary, and image locations will depend on where the object is placed between the focal point and the lens.
ii. Do not agree! Virtual images can be seen but cannot be projected on a screen. A simple demonstration would be to produce a real image and project it on a screen. Then produce a virtual image and show that the image cannot be made to show up on the screen.
419. (a) The left figure rays should all be parallel and pointing downward, with the angle of refraction larger than angle of incidence. The right figure should also show the waves traveling in a more downward direction. All wave fronts should be parallel, and the wavelength between the wave fronts should be larger in medium #2.
(b) i. The sketched waves should bend around the boundary and overlap in circular paths that maintain the same wavelength.
ii. The point source model says that every point on the wave is the source of a new wavelet that propagates outward. For the plane waves approaching the barrier, the sum of all the wavelets produces another plane wave in front of the last one. The barrier blocks the center wavelets. The plane waves on either side of the boundary continue forward, but the wavelets on the end of the blocked wave produce curved waves that propagate inward to fill the central area beyond the boundary. These curved waves will overlap and form constructive interference, where crests meet crests and troughs meet troughs. They will create destructive interference where crests meet troughs.
(c) i. There are several ways to draw this. The interference pattern shown here is marked to show the constructive interference points. An amplitude function could also be used to represent the interference pattern. The key points of the drawing are that it is symmetrical along the central axis and that there is a central maximum. We do not have enough information to calculate the exact locations of the constructive and destructive interference. What is important is that they alternate in an evenly spaced pattern and are marked with C and D as seen in the figure.
ii. The pattern will get wider and spread out from the central maximum. As the distance between the wave fronts is increased, the wavelength gets larger. This means the pattern will spread out as the angle θ gets larger: dsinθ = mλ.
420. (a) i. The ray must turn away from the normal to the right.
ii. The speed of light in air is approximately 3 × 108 m/s. In glass, it travels more slowly. When the ray exits the glass and enters air, the light speeds up. This causes the wavelength to get larger, and the wave turns away from the normal line.
(b) Writing Snell’s law for this situation and lining it up with the equation for a line, we see that if we plot sinθa on the y-axis and sinθg on the x-axis, we should get a straight line with a slope of ng and an intercept of zero:
(c) Equipment: protractor, pencil, ruler, sheet of white paper
1. Place the semicircular prism in the center of the sheet of paper.
2. Trace the outside of the prism with the pencil. Mark a spot in the center of the flat surface of the prism where the laser beam is to exit each time.
3. Direct the laser beam perpendicular through the curved surface of the prism toward the spot in the center of the flat surface. Use the ruler to verify that the beam goes straight through the curved surface of the prism without bending.
4. Trace the ray’s path with the ruler before it enters the prism’s curved surface and after it exits the prism at the mark on the flat side.
5. Repeat for a wide variety of angles, and number each incoming and outgoing ray so you know which one is paired up with which when you measure the angles later.
6. Remove the prism. Mark the normal line at the flat surface. Extend the light rays to the normal line/flat surface intersection.
7. Measure the angles of incidence and refraction for each incoming and outgoing ray.
8. Build a table of data and plot sinθa on the y-axis and sinθg on the x-axis. The slope will be the index of refraction of the glass.
(d) i. Traveling along path A, the light reflects and refracts at the top surface. Some light escapes the top. The rest reflects toward the left surface, where some of the light refracts and escapes. Traveling along path B, the light hits the top surface at an angle larger than the critical angle, and none can escape the top surface. All the light reflects toward the left surface, where some refracts and escapes.
ii. and iii.
iv. Path B will produce the brightest beam, exiting on the left side. None of the path B beam exited the top of the prism. By conservation of energy, there will be more light left to exit the prism on the left side.
Chapter 7: Quantum, Atomic, and Nuclear Physics
421. (A) The speed of light. Relativity says that no matter what, everyone measures the same speed of light.
(B) The speed of light: 3 × 108 m/s.
422. The laws of physics are the same in all inertial reference frames. Thus, the laws of physics are the same for everyone. The speed of light is always 3 × 108 m/s in all inertial reference frames. Thus, everyone measures the same speed of light, even if the source or the observer is moving.
423. We start to see the effects of time dilation and length contraction when moving about one tenth the speed of light and faster. Thus, we usually don’t see the effects of relativity in our everyday lives.
424. A person moving away from the earth in a spaceship with a speed of 0.5 c will not agree with a person on earth about the time it takes to get to the nearest star. Nor will they agree on the distance to the star. Events that happen on the spaceship seem to happen at different times, as measured by the person in the spaceship and the observer on earth.
425. The photoelectric effect is the phenomenon where light shining on an object can cause electrons to be ejected from (knocked off) the object. Electrons are held by the atoms in the material (work function). The light must have an energy at least as large as or higher than this work function for an electron to be ejected. This is the experiment that showed the particle properties of light, because light of too low a frequency cannot eject electrons no matter how bright the light is. Einstein coined the term photon to describe the new particle nature of light. Photons of light have energy that is dependent on their frequency: E = hf.
426. Ultraviolet photons have enough energy to dislodge electrons from a negatively charged object that has excess electrons. Positively charged objects lack electrons. The only way to discharge a positively charged object is to add electrons or remove protons. A UV photon cannot do either of these.
427. Figure A shows photons of light striking potassium, which has a work function of 2.29 eV. This means it takes 2.29 eV of energy to knock an electron off potassium. Photons of light with less than this value will not be able to eject an electron because photons interact with electrons like particles on a one-to-one basis. Note that 700 nm does not have enough energy to produce photoelectrons. The energy of photoelectrons is measured by forcing them to travel through a potential difference that tries to stop them. The stopping potential is increased until none of the electrons can reach the other plate, and the photocurrent drops to zero, as shown in Figure B. The maximum energy of the ejected electrons is Kmax = Uelectic = qΔV = eΔV = hf – Φ work function.
428. Photon energy: The energy of a single particle of light. Photon energy depends only on the frequency of the wave: E = hf.
Stopping potential: This is the electric potential used to stop the photoelectrons to determine their kinetic energy: Kmax = Uelectric = qΔV = eΔV.
Threshold/cutoff frequency: This is the frequency below which photons of light no longer have enough energy to produce photoelectrons (knock electrons off the metal): hfT = Φwork function.
Work function: This is the minimum amount of energy needed to remove an electron from a material. Think of it as how much energy the atom expends to hold on to the outermost electron.
429. There are two phenomena in the photoelectric experiment that support the particle model of light. First, both dim light and very bright (intense) light of the same wavelength and frequency eject electrons with exactly the same maximum kinetic energy. This does not make sense according to the wave model of light, which says that brighter light has a larger amplitude and should eject electrons with more kinetic energy. The particle model of light solves this problem. Dim light is simply a stream of a few photons. Bright light is a stream of many photons. Dim light will eject fewer photoelectrons than bright light because it has fewer photons. However, the ejected electrons from both dim light and bright light have exactly the same kinetic energy. This is because all the photons have the same frequency and, therefore, the same energy.
Second, for each photosensitive material tested, the energy of the ejected photoelectrons was directly related to the frequency of the light striking the surface. Even stranger, below a certain threshold/cutoff frequency, no electrons will be ejected no matter how intense or bright the light is. According to the wave model, it should be possible to use intense but low frequency light to remove electrons from the material. The particle model solves this problem as well. Light consists of particles (photons) of light whose energy is directly proportional to the frequency of the wave: E = hf. Low frequency light cannot eject electrons because it simply does not have the energy to do so. The higher the frequency of light, the more energy the photoelectrons will have: Kmax = hf – Φwork finction.
430. The energy of all electromagnetic (EM) waves depends only on the frequency of the light: E = hf.
432. An electron volt is a unit of energy convenient when in the atomic scale: 1 eV = 1.6 × 10–19 C.
433. (A) At the threshold frequency, the kinetic energy of the photoelectrons is zero:
434. (A) Wave property: Only waves exhibit interference.
(B) Particle property: Light behaves like a particle by colliding with an electron, thus imparting momentum to the electron and knocking it from the atom.
(C) Wave property: Interference is a wave property.
(D) Particle property: Collisions and momentum are particle properties.
(E) Wave property: Diffraction, the bending of waves around boundaries, is a wave property.
(F) Wave property: The Doppler effect is a property of waves.
(G) Particle property: Solar cells, or photovoltaic cells, are a practical application of the photoelectric effect, which demonstrates the particle nature of light.
435. No! Electromagnetic waves/photons of light do not have a net charge and are immune to external electric and magnetic fields.
436. Photons are massless particles that have both energy and momentum.
437. Photon energy: E = hf. Photon momentum: .
Remember that for a wave, v = fλ, and the speed of a light wave is c.
The relationship between photon energy and momentum: .
438. (A) The X-ray will lose energy and momentum. Conservation of momentum must be obeyed in the collision. Since the electron gained momentum in the collision, the photon must lose momentum. Also, the electron gains kinetic energy in the collision. Therefore, the photon must lose energy. Losing momentum and energy means the frequency of the X-ray has decreased: E = hf = pc.
(B) The electron gains both kinetic energy and momentum to the right from the collision with the X-ray.
439. (A) The X-ray will lose energy and momentum. Conservation of momentum must be obeyed in the collision. Since the electron gained momentum in the collision, the photon must lose momentum. Also, the electron gains kinetic energy in the collision. Therefore, the photon must lose energy. Losing momentum and energy means the frequency of the X-ray has decreased: E = hf = pc.
(B) Elastic! Electrons have no internal structure. Therefore, there are no dissipative internal forces to turn the kinetic energy into thermal energy.
Momentum in the x-direction:
Momentum in the y-direction:
440. The Rutherford model of the atom consisted of a nucleus in the center of the atom where all the positive charge resided. This area is approximately 10–14 m in diameter. The diameter of the atom as a whole was approximately 10–10 m in diameter. This means the nucleus was much smaller than the atom itself. The rest of this atomic space around the nucleus was where the tiny elections resided. At the time, nobody had any idea what the electrons were doing in this big open area of the atom.
441. Accelerating charged particles radiate energy in the form of EM waves. Orbiting electrons would lose energy by radiation and crash into the nucleus, destroying the atom. So it cannot be correct.
442. Bohr “fixed” the planetary model of electron orbits by assuming stable orbital locations where the electrons would not radiate energy.
443. To move from one stable state to another required the emission or absorption of a photon of light. Moving up to a higher energy state required the absorption of a photon. Falling to a lower energy state required the emission of a photon. The photon energy always equals the difference in energy of the two stable energy states.
444. The 7-eV photon will be absorbed taking the electron from the ground state to the n = 3 energy level, the second excited state, as shown in the figure. The atom now has 7 eV more energy.
445. The 9 eV photon is not absorbed by an electron in the ground state because there is no stable energy stage at –3 eV. There is no energy change in the atom.
446. The 15-eV photon has enough energy to knock the electron free of the atom with 3 eV of kinetic energy. The atom is ionized.
447. Starting in the n = 4 energy level, the electron can follow any path that eventually takes the electron back to the ground state. All the possible transitions are shown in the figure. Each of these transitions will be accompanied by a photon emission, with energy equal to the difference in energies of the two states.
448. (A) . The frequency is directly related to the energy change in the transition. Therefore, the greater the ΔE, the greater the f of the emission. The frequency ranking from highest to lowest is B = D > A > C.
(B) The wavelength is inversely related to the energy change in the transition. Therefore, the smaller the ΔE, the greater the λ of the emission. The wavelength ranking is opposite that of frequency: C > A > B = D.
(C) Since photons are emitted from the atom during these transitions, the atom is losing energy, as shown in the figure.
449. (A) . The frequency is directly related to the energy change in the transition. Therefore, the greater the ΔE, the greater the f of the absorbed photon. The frequency ranking from highest to lowest is A > B = C.
(B) The wavelength is inversely related to the energy change in the transition. Therefore, the smaller the ΔE, the greater the λ of the absorbed photon. The wavelength ranking is B = C > A.
(C) Since photons are being absorbed during these transitions, the atom is gaining energy.
450. Note that the energy change is –3 eV. This means the atom is losing energy and a photon will be emitted by the atom. The absolute value of the ΔE is taken in the following equation so the answers come out positive.
This is visible! It falls in the visible spectrum (400–700 nm) on the violet end of the spectrum.
The problem does not tell us the starting position of the electron or its initial energy. So let’s just label it at E0. Using the equation we can calculate the change in energy of the electrons in the atom associated with the photons:
The wavelengths of 400 nm and 600 nm give us energies of 3.1 eV and 2.1 eV, respectively.
(B) Yes! The figure shows that when the electron is in the E0 + 3.1 eV energy level, it can fall to the E0 + 2.1 eV energy level. This will produce a 1,240 nm/1.0 eV photon that is not in the visible spectrum, so it will not be seen.
452. The energy of the atom increases during the pumping phase as electrons are moved to higher energy states. The atom loses energy during the lasing phase as the stored energy of the electron in the E2 energy level is released in the form of a photon.
453. Energy gain during the pumping phase is equal to the energy change during the two transitions:
454. The energy of the photon equals the energy difference of the two energy states. This will be a negative number because the atom is ejecting a photon, so it is shown as an absolute value:
455. Ephoton #1 = Ephoton #2 + Ephoton #3
456. Scientists have shown that massless EM waves have particle properties. Louis de Broglie suggested that particles with mass might also have wave properties. This turned out to be true and is one of the foundations of quantum mechanics. The wave nature of particles is named after de Broglie. The de Broglie wavelength of a particle with mass depends on the particle’s momentum (mv): .
457. Both are inverse relationships, as seen in the equation for the de Broglie wavelength: .
458. Shooting small particle, like electrons, through a double slit or single slit will produce an interference pattern, which is a wave property. Directing a beam of electrons at a crystal produces a diffraction pattern similar to that produced by X-rays. Electrons having wave properties explains why Bohr’s stable electron energy states exist (see the answer to question 460). It also explains the electron energy shells that we all learned about in chemistry.
459. Electron beams can produce de Broglie wavelengths on the same order of magnitude and smaller than the atomic crystal spacing. A baseball will generally have a wavelength many magnitudes smaller than the atomic spacing in a crystal. Baseball wavelengths are much too small to produce an interference pattern. (This is the same reason light does not produce interference patterns when passing through two doorways, but sound waves do.) In addition, the baseball itself is larger than the atomic spacing of the crystal, so it is not physically possible for the baseball to interact with a single atom of the crystal at a time.
460. The electron only exists in “orbits” that are whole number multiples of the electron wavelength. These orbits are “stable” because the electron exhibits constructive interference in this location. At other locations, the electron wavelength is either too long or too short to produce constructive interference. The electron cannot exist in these locations because there is destructive interference, and the orbit is not stable.
461. (A) This is an interference pattern. Therefore, electrons are exhibiting wave properties (wave-particle duality).
(B) When the accelerating voltage increases, the electron velocity increases: When electron velocity increases, the de Broglie wavelength of the electron decreases: When wavelength decreases, the pattern gets tighter, and the angle of the maxima decreases: dsinθ = mλ.
(C) The interference pattern will become more spread out as the distance d decreases:
462. A particles wave function describes the wave nature of the particle. It shows us the probability of finding the particle at a particular location. The larger the amplitude of the wave function, positive or negative, the higher likelihood of the particle being found at that location.
463. (A and B)
464. (A) This is a capacitor. The top plate has a higher electric potential. The electric field is directed straight down toward the bottom plate.
(B) Ray C must not have a charge because it is not affected by the electric field. Ray B curves in the direction of the electric field. Therefore, it must be receiving a force from the field and must be positively charged. Ray A must be negative as it receives a force in the opposite direction of the electric field and curves upward.
(C) Ray C could be a neutron or a gamma ray because neither has an electric charge. Ray B could be a proton or an alpha particle. Ray A could be an electron. Ray A curves more tightly, which could be accounted for by an electron having less mass and therefore more acceleration than an alpha particle or a proton as in the case of ray B. However, we do not know the speed of the particles exiting the box. So it is hard to draw any conclusions by comparing the curve radius of ray A and B.
466. The atomic mass unit is a convenient unit to use when dealing with atomic-size particles. It is equal to 1.66 × 10–27 kg = 931 MeV/c2. This is 1/12 of the mass of a carbon-12 nucleus.
467. Note that several versions of the particle symbol are shown in the table. Several particles in the table have different names but are otherwise identical.
468. Atoms consist of a nucleus composed of protons and neutrons. The nucleus is about 10–14 m in diameter. Electrons exist outside the nucleus in a region that is about 10–10 m in diameter.
469. The electrons are held in the atom by the electrostatic attraction between the positive protons in the nucleus and the negative electrons.
470. The protons are repelled by the electric force. Gravity pulls the nucleons together but is too weak to hold the nucleus together. The nuclear strong force is stronger than electrostatic repulsion of the protons. The strong force attracts all the nucleons together, making the nucleus stable.
471. Isotopes are nuclei with the same number of protons (same element) but a different number of neutrons.
472. Not all atoms are stable. The nuclear strong force is a short-range force that holds adjacent nucleons together. As the nucleus grows, electrostatic repulsion of the protons can cause decay, as in uranium. Most isotopes are unstable. There are 15 known isotopes of carbon. Only C-12 and C-13 are stable. C-14 has a half-life of 5,730 years. None of the isotopes of uranium are stable.
473. Isotopes have the same number of protons and electrons. Since the electrons are the same, all isotopes have the same chemical properties and chemical behavior. Isotopes do not have the same nuclear properties because some are stable and others are not.
474. The mass of a nucleus is always less than the sum total of the individual masses of the protons and neutrons that comprise it. The difference in mass between a nucleus and its nucleon parts is the mass defect. Some of the nucleon mass is converted into energy that binds the nucleus together: E = mc2. To break a nucleus apart, energy equal to the mass defect must be added to the nucleus to reproduce the missing mass. The mass defect and binding energy is an “energy” explanation for nuclear stability. The nuclear strong force is a “forces” explanation for why the nucleus stays together.
475. Similarities: Both nuclear binding energy and ionization energy refer to the input energy required to separate the particles from a nucleus or an electron from an atom. To separate a nucleon from a nucleus requires energy input to recreate the missing mass of the mass defect. To separate an electron from an atom, energy input is needed to overcome the attraction of the nucleus and the electron.
Differences: The nuclear binding energy is much larger than the ionization energy.
476. Δm = (2mneutron + 2mproton) − mHe
477. Find the total mass of the 26 protons and 30 neutrons that make up this nucleus:
Find the mass defect by subtracting the mass of the Fe nucleus from the constituent parts:
To find the binding energy, we use Einstein’s famous equation:
478. Conservation of charge: This is the same as conservation of atomic number or Z-number.
Conservation of nucleon number: This is the same as conservation of atomic mass number or A-number.
Conservation of mass/energy: This equivalent to conservation of energy, realizing that mass is equal to the rest energy: E = mc2.
Conservation of momentum: This is just like momentum collisions in mechanics. For example, during alpha decay, the unstable nucleus will shoot out an alpha particle at high speed, while the nucleus will recoil at slow speed due to its larger mass. Note that we can use the atomic mass number (A-number) as the mass of the reactants.
479. Alpha decay: A nucleus is ejected from the unstable nucleus, causing a loss of two protons and two neutrons. The alpha particle leaves with kinetic energy; therefore, the mass of the products are less than that of the original nucleus. The nucleus will recoil in the opposite direction of that of the departing alpha particle.
Beta decay: An electron is ejected from the unstable nucleus, causing a transformation of a neutron into a proton. The beta particle leaves with kinetic energy; therefore, the mass of the products are less than that of the original nucleus. The original nucleus will recoil in the opposite direction of that of the departing beta particle. This recoil will be less than that caused by an alpha particle due to the smaller mass of the beta particle.
Gamma decay: An excited state of a nucleus ejects a gamma ray as it transitions to a lower energy state. Gamma rays are massless and changeless and do not affect the nucleon number, or charge of the nucleus. The gamma ray takes energy away from the nucleus. Therefore, the mass of the nucleus decreases. In theory, the nucleus will recoil when the gamma ray departs, but the recoil will be very small.
480. Nuclear decay occurs when a nucleus is unstable because there is not enough binding energy to hold it together due to an excess number of nucleons. Mass is lost in the process (the mass defect increases), and the products of decay are more stable than the reactants.
481. Mass is lost in the process of decay. The mass defect increases during decay. This makes the products of decay more stable than the original nuclei.
482. Fission: Sometimes when a large nucleus is bombarded by a neutron, the nucleus becomes unstable and splits into two smaller nuclei and multiple neutrons. Both the nuclear charge and the number of nucleons are conserved in the process. The final products are more stable as mass is converted into binding energy. Some of the nuclear mass is also converted into kinetic energy of the reactants. The expelled neutrons can fission more nuclei, causing a cascading reaction (nuclear reactor and nuclear bomb).
Fusion: Given enough initial velocity, two small nuclei can be forced together past electrostatic repulsion to where the nuclear strong force will fuse them together to create a larger nucleus. During the process, all four conservations are obeyed. The final nucleus has less mass, as energy is released like in the sun.
483. (A) Not possible! The number of nucleons is not conserved.
(B) Not possible! Conservation of charge is violated.
485. Sample #1: approximately 3.5 years. Sample #2: approximately 2.5 minutes.
486. Half-life is the time it takes for one half of a radioactive sample to decay into a new element. After two half-lives, a quarter of the original radioactive substance will remain.
487. 1.6 decays/min/g is approximately one eighth of the original activity: (12.6/2/2/2 = 1.57). This is three half-lives. Therefore, the wood is approximately 17,000 years old: 5,730 years × 3 half-lives = 17,190 years.
488. (A) To find the energy released, first find the mass defect:
(B) Calculate the velocity of the alpha particle:
Use conservation of momentum to find the velocity of the polonium nucleus. Since the original velocity of the radon atom is not given, we assume that it was stationary:
(C) The half-life of radon is 3.8 days. Dividing 19 days by the 3.8 days half-life, we see that five half-lives have passed.
(D) Polonium is also radioactive with a 3.10-minute half-life. This is a much shorter half-life than that of radon. Most of the polonium will decay into a new element shortly after being created. So we can’t really answer this question except to say that most of the radon that has decayed into polonium will already have been transmuted into some new element.
AP-Style Multiple-Choice Questions
489. (C) Special relativity tells us that when two observers are moving relative to each other, they will not necessarily agree on length and time. This becomes evident when we get up near the speed of light. We start to notice the effect around 0.1 c and faster. The only constant that all observers will agree on is the speed of light.
490. (D) The difference in radius is small. The radius of curvature for the ions is given by . The radius is directly proportional to the mass and inversely proportional to the charge of the ions: . A different isotope of neon would have a small difference in mass that could account for the small difference in radius. It is possible that the neon could have been singly, doubly, or triply ionized. However, this would change the charge by a whole-number factor. This would cause the different radii to be whole-number multiples of each other. Neon cannot have different numbers of protons without becoming a new element. This is not an interference pattern, so it cannot be formed by wave interference.
491. (C) The particle is most likely to be found at the highest positive/negative amplitude location of Ψ as a function of x. The particle will not be found at locations –15 nm and 0.0 nm.
492. (C) Solve for A and Z. A = 32 and Z = 16. Knowing that alpha particles have two protons and four nucleons, we can divide 32 by 4 or divide 16 by 2 to find out that we need 8 alpha particles to balance the equation.
493. (B and C) Conservation of charge is satisfied because the net charge before and after is zero. Conservation of momentum tells us that the initial momentum of the gamma ray must be equal to the net momentum of the two particles. The gamma ray has x-direction momentum and no y-direction momentum. This appears to be stratified in the figure. The Answer choice D is incorrect; we have to use conservation of mass/energy because most of the gamma ray energy has been converted into mass not just kinetic energy of the two particles.
494. (C) Adding the excitation energy of 100 eV – 115 eV to the ground state energy, we get the electron energy range of (-22.4 eV) – (-7.4 eV). This will take the electrons of the gas from the ground state to the n = 3 and 4 energy levels. The highest frequency photon occurs when the electron jumps down in energy by the largest amount: E = hf. Therefore, the transition from n = 3 to n = 2 is the answer. Answer choices A and D are not possible because the electrons were not excited to the n = 5 level.
495. (A and C) A higher work function would account for lower energy electron emissions from A: Kmax = hf − Φ. Placing the plate closer to the light source would increase the number of photons captured by the plate, but it will not increase the energy of the photons striking the plate. Thus, it would produce more electrons, but the energy would remain the same for each electron.
496. (B and C) The discreet energy levels of electrons in an atom can be accounted for by the wave nature of the electron and constructive and destructive interference. The diffraction is a wave property. The photoelectric effect demonstrates the particle nature of light.
497. (D) The electrostatic repulsion of the protons is much stronger than the gravitational attraction of the nucleons. Thus, we need a stronger force to hold the nucleus together. Scientists call this the nuclear strong force.
498. (B) The energy of the gamma ray comes from the mass that is lost in the reaction. Adding the mass of the reactants, subtracting the mass of the products, and converting into energy via E = mc2 gives us the mass defect that was converted into the energy of the gamma ray. This is an application of conservation of mass/energy.
AP-Style Free-Response Questions
499. (a) Helium has two protons and four neutrons. Completing the nuclear equation with conservation of nucleon number and conservation of charge, we get . Therefore, lithium has three protons and two neutrons.
(b) Gravity pulls the nucleons together, but it is very weak. The protons are repelled by the electric force. The nuclear strong force is stronger than all of these forces and attracts all the nucleons together, making the nucleus stable.
(c) The kinetic energy comes from the mass that is lost in the reaction that is converted into energy:
(d) Conservation of momentum applies. The lithium was at rest. Therefore, the helium and proton must move off in opposite directions. The helium has more mass and will move more slowly than the lighter proton.
(e) First convert to standard units:
Proton: (1.0073 u)(1.66 × 10–27 kg/u) = 1.672 × 10–27 kg
Helium: 4.0012 u = 6.642 × 10–27 kg.
Converting energy to joules: (2475 × 103 eV)(1.6 × 10–19 J/eV) = 3.96 × 10–13 J
Use conservation of mass/energy:
Insert the values into the equation, and use 3.0 × 108 m/s for the speed of light:
Lithium mass = 8.318 × 10–27 kg
Note that only 0.004 × 10–27 kg of mass is needed to create the 2,475 MeV of energy in this reaction.
(f) They will need to manufacture it. The half-life is very short, and this isotope will decay away into other atoms in less than a fraction of a second.
500. (a) First convert the wavelengths of light into energy in electron volts:
248 nm 5 eV, 400 nm 3.1 eV, 650 nm 1.9 eV
With this information, we can build the energy level diagram. Remember that absorbing photons causes the electron to jump upward. Starting in the ground state of –8 eV and adding a 5-eV absorbed photon, we see that there is an energy level at –3 eV. To produce the 400-nm and 600-nm light, there must be an intermediate level between the ground state and the –3-eV level. There are two acceptable ways to draw this intermediate energy level, as shown in the two figures, depending on whether the 650-nm photon is emitted first or second. Be sure to label the energies of each level.
(b) No, it will not eject electrons. The 650-nm photon has even less energy than the 400-nm photon and will be below the threshold needed to eject and electron:
(c) i. We need to convert electron energy into joules of kinetic energy:
(2.71 eV)(1.6 × 10–19 J/eV) = 4.34 × 10–19 J
Calculate the velocity of the electron. Remember that the mass of an electron is 9.11 × 10–31 kg:
Calculate the de Broglie wavelength. Remember that h = 6.63 × 10–34 J · s:
ii. The equation shows us that the ratio needs to be less than 1 for an interference pattern to form. Any larger ratio will cause the angle θ to be larger than 90 degrees, and no pattern will form. This means the wavelength must be smaller than the spacing, but not too small or the angle will be so small that the pattern will be too small to see. The electrons will form an interference pattern, as the waves are on the same order of magnitude and smaller than the opening spacing. As a side note, none of the light emitted from the gas will form an interference pattern because the wavelengths are much larger than the opening spacing.