## MCAT Physics and Math Review

## Chapter 9: Atomic and Nuclear Phenomena

### Practice Questions

1. If the work function of a metal is 6.622 × 10^{–20} J and a ray of electromagnetic radiation with a frequency of 1.0 × 10^{14} Hz is incident on the metal, what will be the speed of the electrons ejected from the metal? (Note: *h* = 6.626 × 10^{−34} J**·**s and *m*_{e}^{−} = 9.1 × 10^{−31} kg)

1.

2.

3.

4.

2. What is the wavelength of a photon that causes an electron to be emitted from a metal with a kinetic energy of 50 J? (Note: The work function of the metal is 16 J, and *h* = 6.626 × 10^{−34} J**·**s)

1. 1.0 × 10^{−34} m

2. 3.0 × 10^{−27} m

3. 3.0 × 10^{−26} m

4. 1.0 × 10^{35} m

3. Which of the following statements is inconsistent with the Bohr model of the atom?

1. Energy levels of the electron are stable and discrete.

2. An electron emits or absorbs radiation only when making a transition from one energy level to another.

3. To jump from a lower energy to a higher energy, an electron must absorb a photon of precisely the right frequency such that the photon’s energy equals the energy difference between the two orbits.

4. To jump from a higher energy to a lower energy, an electron absorbs a photon of a frequency such that the photon’s energy is exactly the energy difference between the two orbits.

4. When a hydrogen atom electron falls to the ground state from the *n* = 2 state, 10.2 eV of energy is emitted. What is the wavelength of this radiation? (Note: 1 eV = 1.60 × 10^{−19} J, and *h* = 6.626 × 10^{−34} J**·**s.)

1. 5.76 × 10^{−9} m

2. 1.22 × 10^{−7} m

3. 3.45 × 10^{−7} m

4. 2.5 × 10^{15} m

5. The figure below illustrates an electron with initial energy of –10 eV moving from point A to point B. What change accompanies the movement of the electron?

1. Absorption of a photon

2. Emission of a photon

3. Decrease in the atom’s work function

4. Increase in the atom’s total energy

6. Ultraviolet light is more likely to induce a current in a metal than visible light. This is because photons of ultraviolet light:

1. have a longer wavelength.

2. have a higher velocity.

3. are not visible.

4. have a higher energy.

7. All of the following statements about the photoelectric effect are true EXCEPT:

1. the intensity of the light beam does not affect the photocurrent.

2. the kinetic energies of the emitted electrons do not depend on the light intensity.

3. a weak beam of light of frequency greater than the threshold frequency yields more current than an intense beam of light of frequency lower than the threshold frequency.

4. for light of a given frequency, the kinetic energy of emitted electrons increases as the value of the work function decreases.

8. What is the binding energy of the argon-40 isotope in MeV? (Note: *m*_{proton} = 1.0073 amu, *m*_{neutron} = 1.0087 amu, *m*_{Ar-40 nucleus} = 39.9132 amu, )

1. 0.4096 MeV

2. 40.3228 MeV

3. 381.7 MeV

4. 643.8 MeV

9. Which of the following correctly identifies the following process?

1. *β*^{−} decay

2. *β*^{+} decay

3. *e*^{−} capture

4. *γ* decay

10.Consider the following fission reaction.

The masses of the species involved are given in atomic mass units below each species, and 1 amu can create 932 MeV of energy. What is the energy liberated due to transformation of mass into energy during this reaction?

1. 0.003 MeV

2. 1.4 MeV

3. 2.8 MeV

4. 5.6 MeV

11.Element X is radioactive and decays via *α* decay with a half-life of four days. If 12.5 percent of an original sample of element X remains after *n* days, what is the value of *n*?

1. 4

2. 8

3. 12

4. 16

12.A graph of an exponential decay process is created. The *y*-axis is the natural logarithm of the ratio of the number of intact nuclei at a given time to the number of intact nuclei at time *t* = 0. The *x*-axis is time. What does the slope of such a graph represent?

1. *λ*

2. *−* *λ*

3. *e*^{−}*λt*

4.

13.A certain carbon nucleus dissociates completely into *α* particles. How many particles are formed?

1. 1

2. 2

3. 3

4. 4

14.The half-life of carbon-14 is approximately 5730 years, while the half-life of carbon-12 is essentially infinite. If the ratio of carbon-14 to carbon-12 in a certain sample is 25% less than the normal ratio in nature, how old is the sample?

1. Less than 5730 years

2. Approximately 5730 years

3. Significantly greater than 5730 years, but less than 11460 years

4. Approximately 11460 years

15.A nuclide undergoes two alpha decays, two positron decays, and two gamma decays. What is the difference between the atomic number of the parent nuclide and the atomic number of the daughter nuclide?

1. 0

2. 2

3. 4

4. 6

PRACTICE QUESTIONS

### Answers and Explanations

1. **C**

To determine the speed of the electrons ejected, we must first calculate their kinetic energy:

Using the formula for the kinetic energy, we can now calculate the speed of the ejected electrons:

Notice the wide range in the exponents for the answer choices. While the math in this question may seem complex, this allows us to round significantly.

2. **B**

To determine the wavelength of the light ray, first calculate its frequency from the photoelectric effect equation:

Next, determine the wavelength of the incident ray of light by relating the frequency to the speed of light:

3. **D**

The Bohr model is based on a set of postulates originally put forward to discuss the behavior of electrons in hydrogen. In summary, these postulates state that the energy levels of the electron are stable and discrete, and they correspond to specific orbits, eliminating **choice (A)**. They also state that an electron emits or absorbs radiation only when making a transition from one energy level to another, eliminating **choice (B)**. Specifically, when an electron jumps from a lower-energy orbit to a higher-energy one, it must absorb a photon of light of precisely the right frequency such that the photon’s energy equals the energy difference between the two orbits, eliminating **choice (C)**. When falling from a higher-energy orbit to a lower-energy one, an electron emits a photon of light with a frequency that corresponds to the energy difference between the two orbits, This is the opposite of **choice (D)**, which makes it the right answer.

4. **B**

To solve this question correctly, one must be careful with the units. First, convert 10.2 eV to joules:

Next, to determine the wavelength of the radiation, first find the frequency:

Lastly, from the wave equation c = *f* *λ*, we can calculate the wavelength of the radiation:

5. **B**

The electron moves from a higher energy level to a lower energy level; this can only occur if the extra energy is dissipated through the emission of a photon. If the electron moved from B to A, it would absorb a photon and increase the atom’s total energy; however, the opposite is occurring, so **choices (A)** and **(D)** can be eliminated. The work function is the amount of energy required to eject an electron from a material; when moving from A to B, the electrical potential energy of the atom decreases, meaning that more energy will be required to free the electron from the atom, eliminating **choice (C)**.

6. **D**

The photoelectric effect occurs when a photon of sufficiently high energy strikes an atom with a sufficiently low work function. This means that a photon with higher energy is more likely to produce the effect. Because ultraviolet light has a higher frequency and lower wavelength than visible light, it also carries more energy according to the equation *E* = *hf*. All light travels at the speed of light, eliminating **choice (B)**. As mentioned earlier, ultraviolet light has a shorter wavelength than visible light, eliminating **choice (A)**. The visibility of a wave plays no role in its ability to cause the photoelectric effect, eliminating **choice (C)**.

7. **A**

The greater the intensity, the greater the number of incident photons and, therefore, the greater the number of photoelectrons that will be ejected from the metal surface (provided that the frequency of the light remains above the threshold). This means a larger current. Remember that the frequency of the light (assuming it is above the threshold frequency) will determine the kinetic energy of the ejected electrons; the intensity of the light determines the number of electrons ejected per time (the current).

8. **C**

To determine the binding energy, we must first determine the mass defect. The mass defect is simply the masses of each of the protons and neutrons in the unbound state added together minus the mass of the formed argon-40 nucleus (which contains 18 protons and 40 − 18 = 22neutrons):

The binding energy can then be determined from this mass defect:

The closest answer is **choice (C)**.

9. **C**

This process can be described as electron capture. Certain unstable radionuclides are capable of capturing an inner electron that combines with a proton to form a neutron. The atomic number becomes one less than the original, but the mass number remains the same. Electron capture is a relatively rare process and can be thought of as the reverse of *β*^{−} decay. Notice that the equation is similar to that of *β*^{+} decay but not identical because a particle is absorbed, not emitted.

10.**C**

This problem presents a reaction and asks for the energy liberated due to transformation of mass into energy. To convert mass into energy, we are told that 1 amu can be converted into 932 MeV of energy. All we need to do now is calculate how much mass, in amu, is converted in the reaction. Because we are given the atomic mass for each of the elements in the reaction, this is simply a matter of balancing the equation:

This is the amount of mass that has been converted into energy. To obtain energy from mass, we have to multiply by the conversion factor (1 amu = 932 MeV):

*E* = 0.003 × 932 ≈ 0.003 × 900 = 2.7 MeV

11.**C**

Because the half-life of element X is four days, 50 percent of an original sample remains after four days, 25 percent remains after eight days, and 12.5 percent remains after 12 days. Therefore, *n* = 12 days. Another approach is to set where *x* is the number of half-lives that have elapsed. Solving for *x* gives *x* = 3. Thus, 3 half-lives have elapsed, and because each half-life is four days, we know that *n* = 12 days.

12.**B**

The expression *n* = *n*_{0}*e*^{−}*λt* is equivalent to Taking the natural logarithm of both sides, we get:

From this expression, it is clear that plotting ln *vs*. *t* will give a straight line with a slope of −*λ*.

13.**C**

A typical carbon nucleus contains 6 protons and 6 neutrons. An *α* particle contains 2 protons and 2 neutrons. Therefore, one carbon nucleus can dissociate into particles.

14.**A**

Because the half-life of carbon-12 is essentially infinite, a 25 percent decrease in the ratio of carbon-14 to carbon-12 means the same as a 25 percent decrease in the amount of carbon-14. If less than half of the carbon-14 has deteriorated, then less than one half-life has elapsed. Therefore, the sample is less than 5730 years old. Be careful with the wording here—the question states that the ratio is 25% *less* than the ratio in nature, not 25% *of* the ratio in nature, which would correspond to **choice (D)**.

15.**D**

In alpha decay, an element loses two protons. In positron decay, a proton is converted into a neutron. Gamma decay has no impact on the atomic number of the nuclide. Therefore, two alpha decays and two positron decays will yield a daughter nuclide with six fewer protons than the parent nuclide.