5 Steps to a 5: AP Chemistry 2017 (2016)
Review the Knowledge You Need to Score High
IN THIS CHAPTER
Summary: Radioactivity, the spontaneous decay of an unstable isotope to a more stable one, was first discovered by Henri Becquerel in 1896. Marie Curie and her husband expanded on his work and developed most of the concepts that are used today.
Throughout this book, you have been studying traditional chemistry and chemical reactions. This has involved the transfer or sharing of electrons from the electron clouds, especially the valence electrons. Little has been said up to this point regarding the nucleus. Now we are going to shift our attention to nuclear reactions and, for the most part, ignore the electron clouds.
Keywords and Equations
No specific nuclear equations are provided, but review first-order equations in the Kinetics chapter.
Balancing Nuclear Reactions
Most nuclear reactions involve breaking apart the nucleus into two or more different elements or subatomic particles. If all but one of the particles is known, the unknown particle can be determined by balancing the nuclear equation. When chemical equations are balanced, coefficients are added to ensure that there are the same number of each type of atom on both sides of the reaction arrow. To balance nuclear equations, we ensure that there is the same sum of both mass numbers and atomic numbers on both the left and right of the reaction arrow. Recall that a specific isotope of an element can be represented by the following symbolization:
In this symbolization A is the mass number (sum of protons and neutrons), Z is the atomic number (number of protons), and X is the element symbol (from the periodic table). In balancing nuclear reactions, make sure the sum of all A values on the left of the arrow equals the sum of all A values to the right of the arrow. The same will be true of the sums of the Z values. Knowing that these sums must be equal allows one to predict the mass and atomic number of an unknown particle if all the others are known.
Consider the transmutation —creation of one element from another—of Cl-35. This isotope of chlorine is bombarded by a neutron and H-1 is created, along with an isotope of a different element. First, a partial nuclear equation is written:
The sum of the mass numbers on the left of the equation is 36 = (35 + 1) and on the right is 1 + x . The mass number of the unknown isotope must be 35. The sum of the atomic numbers on the left is 17 = (17 + 0), and 1 + y on the right. The atomic number of the unknown must then be 16. This atomic number identifies the element as sulfur, so a complete nuclear equation can be written:
Sulfur-35 does not occur in nature; it is an artificially produced isotope.
Natural Radioactive Decay Modes
Three common types of radioactive decay are observed in nature, and two others are occasionally observed.
An alpha particle is a helium nucleus with two protons and two neutrons. It is represented as: or α . As this particle is expelled from the nucleus of the radioisotope that is undergoing decay, it has no electrons and thus has a 2+ charge. However, it quickly acquires two electrons from its surroundings to form the neutral atom. Most commonly, the alpha particle is shown as the neutral particle and not the cation.
Radon-222 undergoes alpha decay according to the following equation:
Notice that in going from Rn-222 to Po-218, the atomic number has decreased by 2 and the mass number by 4.
A beta particle is an electron and can be represented as either or e. This electron comes from the nucleus, not the electron cloud, and results from the conversion of a neutron into a proton and an electron: .
Nickel-63 will undergo beta decay according to the following equation:
Notice that the atomic number has increased by 1 in going from Ni-63 to Cu-63, but the mass number has remained unchanged.
Gamma emission is the giving off of high-energy, short-wavelength photons similar to X-rays. This radiation is commonly represented as γ . Gamma emission commonly accompanies most other types of radioactive decay, but is often not shown in the balanced nuclear equation because it has neither appreciable mass nor charge.
Alpha, beta, and gamma emissions are the most common types of natural decay mode, but positron emission and electron capture are also observed occasionally.
A positron is essentially an electron that has a positive charge instead of a negative one. It is represented as or . Positron emission results from the conversion of a proton to a neutron and a positron: . It is observed in the decay of some natural radioactive isotopes, such as K-40: .
The four decay modes described above all involve the emission or giving off a particle; electron capture is the capturing of an electron from the energy level closest to the nucleus (1s) by a proton in the nucleus. This creates a neutron: . Electron capture leaves a vacancy in the 1s energy level, and an electron from a higher energy level drops down to fill this vacancy. A cascading effect occurs as the electrons shift downward and, as they do so, energy is released. This energy falls in the X-ray part of the electromagnetic spectrum. These X-rays give scientists a clue that electron capture has taken place.
Polonium-204 undergoes electron capture: . Notice that the atomic number has decreased by 1, but the mass number has remained the same. Remember that electron capture is the only decay mode that involves adding a particle to the left side of the reaction arrow.
Predicting whether a particular isotope is stable and what type of decay mode it might undergo can be tricky. All isotopes containing 84 or more protons are unstable and will undergo nuclear decay. For these large, massive isotopes, alpha decay is observed most commonly. Alpha decay gets rid of four units of mass and two units of charge, thus helping to relieve the repulsive stress found in these nuclei. For other isotopes, with atomic numbers less than 84, stability is best predicted by the use of the neutron-to-proton (n/p) ratio.
If one plots the number of neutrons versus the number of protons for the known stable isotopes, the nuclear belt of stability is formed. At the low end of this belt of stability (Z < 20), the n/p ratio is 1. At the high end (Z ≈ 80), the n/p ratio is about 1.5. One can then use the n/p ratio of the isotope under question to predict whether or not it will be stable. If it is unstable, the isotope will utilize a decay mode that will bring it back onto the belt of stability.
For example, consider Ne-18. It has 10 p and 8 n, giving an n/p ratio of 0.8. That is less than 1, so the isotope is unstable. This isotope is neutron-poor, meaning it doesn’t have enough neutrons (or has too many protons) to be stable. Decay modes that increase the number of neutrons, decrease the number of protons, or both would be favored. Both positron emission and electron capture accomplish this by converting a proton into a neutron. As a general rule, positron emission occurs with lighter isotopes and electron capture with heavier isotopes.
Isotopes that are neutron-rich, that have too many neutrons or not enough protons, lie above the belt of stability and tend to undergo beta emission because that decay mode converts a neutron into a proton.
A particular isotope may undergo a series of nuclear decays until finally a stable isotope is formed. For example, radioactive U-238 decays to stable Pb-206 in 14 steps, a majority of which are alpha emissions, as one might predict.
Nuclear Decay Calculations
A radioactive isotope may be unstable, but it is impossible to predict when a certain atom will decay. However, if a statistically large enough sample is examined, some trends become obvious. The radioactive decay follows first-order kinetics (see Chapter 14 on kinetics for a more in-depth discussion of first-order reactions and equations). If the number of radioactive atoms in a sample is monitored, it can be determined that it takes a certain amount of time for half the sample to decay; it takes the same amount of time for half the remaining sample to decay; and so on. The amount of time it takes for half the sample to decay is called the half-life of the isotope and is given the symbol t1/2 . The table below shows the percentage of radioactive isotope remaining versus half-life.
As a general rule, the amount of radioactivity at the end of 10 half-lives drops below the level of detection and the sample is said to be “safe.”
Half-lives may be very short, 4.2 × 10−6 seconds for Po-213, or very long, 4.5 × 109 years for U-238. The long half-lives of some waste products is a major problem with nuclear fission reactors. Remember, it takes 10 half-lives for the sample to be safe.
If only multiples of half-lives are considered, the calculations are very straightforward. For example, I-131 is used in the treatment of thyroid cancer and has a t 1/2 of 8 days. How long would it take to decay to 25% of its original amount? Looking at the chart, you see that 25% decay would occur at two half-lives or 16 days. However, since radioactive decay is not a linear process, you cannot use the chart to predict how much would still be radioactive at the end of 12 days or at some time (or amount) that is not associated with a multiple of a half-life. To solve these types of problems, one must use the mathematical relationships associated with first-order kinetics that were presented in the Kinetics chapter. In general, two equations are used:
In these equations, the ln is the natural logarithm; A t is the amount of isotope radioactive at some time t ; A o is the amount initially radioactive; and k is the rate constant for the decay. If you know initial and final amounts and are looking for the half-life, you would use equation (1) to solve for the rate constant and then use equation (2) to solve for t 1/2 .
For example: What is the half-life of a radioisotope that takes 15 min to decay to 90% of its original activity?
Using equation (1):
If one knows the half-life and amount remaining radioactive, equation (2) can be used to calculate the rate constant k and equation (1) can then be used to solve for the time. This is the basis of C-14 dating, which is used to determine the age of objects that were once alive.
For example, suppose a wooden tool is discovered and its C-14 activity is determined to have decreased to 65% of the original. How old is the object?
The half-life of C-14 is 5,730 yr. Substituting this into equation (2):
Substituting this rate constant into equation (1):
Whenever a nuclear decay or reaction takes place, energy is released. This energy may be in the form of heat and light, gamma radiation, or kinetic energy of the expelled particle and recoil of the remaining particle. This energy results from the conversion of a very small amount of matter into energy. (Remember that in nuclear reactions there is no conservation of matter, as in ordinary chemical reactions.) The amount of energy that is produced can be calculated by using Einstein’s equation E = mc 2 , where E is the energy produced, m is the mass converted into energy (the mass defect), and c is the speed of light. The amount of matter that is converted into energy is normally very small, but when it is multiplied by the speed of light (a very large number) squared, the amount of energy produced is very large.
For example: When 1 mol of U-238 decays to Th-234, 5 × 10−6 kg of matter is converted to energy (the mass defect). To calculate the amount of energy released:
If the mass is in kilograms, the answer will be in joules.
Common Mistakes to Avoid
- Make sure your answer is reasonable. Don’t just write down the answer from your calculator.
- Make sure your units cancel in your calculations, leaving the unit you want.
- Make sure that in alpha, beta, gamma, and positron emissions the particle being emitted is on the right-hand side of the reaction arrow. In electron capture, the electron should be on the left side of the arrow.
- In half-life problems, don’t omit the minus sign. Watch your units.
- In half-life problems, be sure to use the amount of isotope still radioactive asNt and not the amount decayed.
Use these questions to review the content of this chapter and practice for the AP Chemistry exam. Below are 6 multiple-choice questions similar to what you will encounter in Section I of the AP Chemistry exam. To make these questions an even more authentic practice for the actual exam, time yourself following the instructions provided.
Answer the following questions in 10 minutes. You may not use a calculator. You may use the periodic table and the equation sheet at the back of this book.
1 . When decays, it emits 2 α particles, then a β particle, followed by an α particle. The resulting nucleus is:
2 . The formation of from occurs by:
(A) electron capture
(B) α decay
(C) β decay
(D) positron decay
3 . Which of the following lists the types of radiation in the correct order of increasing penetrating power?
(A) α , γ , β
(B) β , α , γ
(C) α , β , γ
(D) β , γ , α
4 . Which of the following statements is correct concerning β particles?
(A) They are electrons, with a mass number of zero and a charge of –1.
(B) They have a mass number of zero, a charge of –1, and are less penetrating than α particles.
(C) They are electrons with a charge of +1 and are less penetrating than α particles.
(D) They have a mass number of zero and a charge of +1.
5 . An atom of undergoes radioactive decay by α emission. What is the product nuclide?
6 . If 75% of a sample of pure decays in 24.6 years, what is the half-life of ?
(A) 24.6 yr
(B) 18.4 yr
(C) 12.3 yr
(D) 6.15 yr
Answers and Explanations for the Multiple-Choice Questions
1 . D —The mass should be 226 – (4 + 4 + 0 + 4) = 214. The atomic number should be 88 – (2 + 2 – 1 + 2) = 83.
2 . B —Mass difference = 234 – 230 = 4, and atomic number difference = 92 – 90 = 2. These correspond to an α particle.
3 . C —Alpha particles are the least penetrating, and gamma rays are the most penetrating.
4 . A —In nuclear reactions, the mass of a β particle is treated as 0, with a charge of –1. Electrons and β particles are the same thing.
5 . B —Mass number = 238 – 4 = 234, and atomic number = 92 – 2 = 90.
6 . C —After one half-life, 50% would remain. After another half-life this would be reduced by one-half to 25%. The total amount decayed is 75%. Thus, 24.6 years must be two half-lives of 12.3 years each.
- Know the five naturally occurring decay modes:
- Alpha emission, in which a helium nucleus, , is emitted from the nucleus.
- Beta emission, in which an electron, , is emitted from the nucleus. This is due to the conversion of a neutron into a proton plus the beta particle.
- Gamma emission, in which high-energy electromagnetic radiation is emitted from the nucleus. This commonly accompanies the other types of radioactive decay. It is due to the conversion of a small amount of matter into energy.
- Positron emission, in which a positron, e , a particle having the same mass as an electron but a positive charge, is emitted from the nucleus. This is due to a proton converting into a neutron and the positron.
- Electron capture, in which an inner-shell electron is captured by a proton in the nucleus with the formation of a neutron. X-rays are emitted as the electrons cascade down to fill the vacancy in the lower energy level.
⊠ Know that nuclear stability is best related to the neutron-to-proton ratio (n/p), which starts at about 1/1 for light isotopes and ends at about 1.5/1 for heavier isotopes with atomic numbers up to 83. All isotopes of atomic number greater than 84 are unstable and will commonly undergo alpha decay. Below atomic number 84, neutron-poor isotopes will probably undergo positron emission or electron capture, while neutron-rich isotopes will probably undergo beta emission.
⊠ Know that the half-life, t 1/2 , of a radioactive isotope is the amount of time it takes for one-half of the sample to decay. Know how to use the appropriate equations to calculate amounts of an isotope remaining at any given time, or use similar data to calculate the half-life of an isotope.
⊠ Know how to use Einstein’s equation E = mc 2 to calculate the amount of energy produced from a mass defect (the amount of matter that was converted into energy).