MCAT Physics and Math Review
Chapter 9: Atomic and Nuclear Phenomena
9.4 Nuclear Reactions
Nuclear reactions, such as fusion, fission, and radioactive decay, involve either combining or splitting the nuclei of atoms. Because the binding energy per nucleon is greatest for intermediate-sized atoms (that is, intermediate-sized atoms are most stable), when small atoms combine or large atoms split, a great amount of energy is released.
When written in isotopic notation, elements are preceded by their atomic number as a subscript and mass number as a superscript . The atomic number (Z) corresponds to the number of protons in the nucleus; the mass number (A) corresponds to the number of protons plus neutrons. When balancing nuclear equations, it is important to balance the number of nucleons on both sizes by balancing the atomic numbers and mass numbers.
Fusion occurs when small nuclei combine to form a larger nucleus. As an example, many stars (including the Sun) power themselves by fusing four hydrogen nuclei to make one helium nucleus as shown in Figure 9.4. By this method, the sun produces 3.85 × 1026 joules per second (385 yottawatts), which accounts for the mass defect that arises from the formation of helium nuclei from hydrogen nuclei. Here on Earth, fusion power plants—which are far less common than fission power plants—generate energy from deuterium and lithium nuclei.
Figure 9.4 Hydrogen Fusion in the Sun, Creating Helium Nuclei
Fission is a process by which a large nucleus splits into smaller nuclei. Spontaneous fission rarely occurs. However, through the absorption of a low-energy neutron, fission can be induced in certain nuclei. Of special interest are those fission reactions that release more neutrons because these other neutrons will cause a chain reaction in which other nearby atoms can undergo fission. This in turn releases more neutrons, continuing the chain reaction. Such induced fission reactions power most commercial nuclear power plants.
A fission reaction occurs when uranium-235 (U-235) absorbs a low-energy neutron, briefly forming an excited state of U-236, which then splits into xenon-140, strontium-94, and more neutrons. In isotopic notation form, the unbalanced reaction is:
When balanced, how many neutrons are produced in the last reaction?
By treating each arrow as an equals sign, the problem is simply asking us to balance the last equation. The mass numbers (A) on either side of each arrow must be equal. This is an application of nucleon number conservation, which says that the total number of neutrons plus protons remains the same, even if neutrons are converted to protons and vice-versa, as they are in some decays. Because 235 + 1 = 236, the first arrow is indeed balanced. Looking at the atomic numbers, the number of protons are balanced throughout (92 + 0 = 92 = 54 + 38 + 0). To find the number of neutrons, determine how many nucleons remain after accounting for xenon-140 and strontium-94:
236 − (140 + 94) = 236 − 234 = 2 nucleons
Because the protons are balanced, these remaining nucleons are both neutrons. Therefore, two neutrons are produced in this reaction. These neutrons are free to go on and be absorbed by more U-235 and cause more fission reactions. Note that it was not actually necessary to know that the intermediate high-energy state was formed.
Radioactive decay is a naturally occurring spontaneous decay of certain nuclei accompanied by the emission of specific particles. On the MCAT, you should be prepared to answer three general types of radioactive decay problems:
1. The integer arithmetic of particle and isotope species
2. Radioactive half-life problems
3. The use of exponential decay curves and decay constants
Isotope Decay Arithmetic and Nucleon Conservation
Let the letters X and Y represent nuclear isotopes. When the parent nucleus X undergoes nuclear decay to form daughter nucleus Y, the balanced reaction is:
When balancing nuclear reactions, the sum of the atomic numbers must be the same on both sides of the equation, and the sum of the mass numbers must be the same on both sides as well.
Whenever you approach radioactive decay problems on the MCAT, start by balancing the number of protons (the atomic numbers). Often, wrong answer choices will simply have an error in the number of protons and can be eliminated before even checking the mass numbers.
Alpha decay is the emission of an α-particle, which is a nucleus that consists of two protons, two neutrons, and zero electrons. The alpha particle is very massive compared to a beta particle and carries double the charge. Alpha particles interact with matter very easily; hence, they do not penetrate shielding (such as lead sheets) very extensively.
Alpha particles do not have any electrons, so they carry a charge of +2.
The emission of an α-particle means that the atomic number of the daughter nucleus will be two less than that of the parent nucleus, and the mass number will be four less. This can be expressed in the balanced equation:
Suppose a parent nucleus X alpha decays as follows:
What are the mass number and atomic number of the daughter isotope Y?
To solve this question, we simply need to balance the atomic numbers and mass numbers:
While it is not necessary to identify the elements to answer the question, answers on the MCAT are usually given with the element’s symbol. Y must be thorium (Th) because its atomic number is 90. Therefore, the daughter nucleus is .
Beta decay is the emission of a β-particle, which is an electron and is given the symbol e− or β−. Electrons do not reside in the nucleus, but they are emitted by the nucleus when a neutron decays into a proton, a β-particle, and an antineutrino ( ) Because an electron is singly charged and 1836 times lighter than a proton, the beta radiation from radioactive decay is more penetrating than alpha radiation. In some cases of induced decay (positron emission), a positron is released, which has the mass of an electron but carries a positive charge. The positron is given the symbol e+or β+. A neutrino (ν) is emitted in positron decay, as well. Note that neutrinos and antineutrinos are not tested on the MCAT, and are therefore omitted in subsequent discussion.
In both types of beta decay, there needs to be conservation of charge. If a negative charge (β−) is created, a neutron is converted into a proton to maintain charge. Conversely, if a positive charge (β+) is created, a proton is converted into a neutron to maintain charge. Remember that negative beta decay creates a negative β-particle and positive beta decay creates a positive β-particle.
During β− decay, a neutron is converted into a proton and a β−-particle (Z = −1, A = 0) is emitted. Hence, the atomic number of the daughter nucleus will be one higher than that of the parent nucleus, and the mass number will not change. This can be expressed in the balanced equation:
During β+ decay, a proton is converted into a neutron and a β+-particle (Z = +1, A = 0) is emitted. Hence, the atomic number of the daughter nucleus will be one lower than that of the parent nucleus, and the mass number will not change. This can be expressed in the balanced equation:
Suppose a promethium-146 nucleus beta-decays as follows:
What are the mass number and atomic number of the daughter isotope Y?
Again, balance the atomic numbers and mass numbers:
Y must be samarium (Sm) because its atomic number is 62. Therefore, the daughter nucleus is .
Gamma decay is the emission of γ-rays, which are high-energy (high-frequency) photons. They carry no charge and simply lower the energy of the parent nucleus without changing the mass number or the atomic number. The high-energy state of the parent nucleus may be represented by an asterisk.
Gamma decay questions are the easiest on the MCAT. No changes occur in the mass number or atomic number; only a γ-ray is emitted.
This can be expressed in the balanced equation:
Suppose an excited parent isotope gamma decays to which then undergoes positron emission to form which in turn alpha decays to If Z is americium-241, what is
Because the final daughter nucleus is given, it will be necessary to work backwards through the reactions. The last reaction is the following α decay:
The atomic number of the parent nucleus must be 97, and the mass number is 245. This is berkelium-245. The preceding reaction is the following positron emission:
The atomic number of the parent nucleus must be 98, and the mass number is 245. This is californium-245. Finally, the preceding reaction is the following gamma decay:
The atomic number of the parent nucleus must be 98, and the mass number is 245. This is a higher-energy form of californium-245:
Certain unstable radionuclides are capable of capturing an inner electron that combines with a proton to form a neutron, while releasing a neutrino. The atomic number is now one less than the original but the mass number remains the same. Electron capture is a rare process that is perhaps best thought of as the reverse of β− decay:
In a sample of radioactive particles, the half-life of the sample is the time it takes for half of the sample to decay. In each subsequent half-life, one-half of the remaining sample decays so that the remaining amount asymptotically approaches zero.
Half-life problems are common on the MCAT. Make sure you write them out; it’s easy to lose your place when doing them in your head.
If the half-life of a certain isotope is 4 years, what fraction of a sample of that isotope will remain after 12 years?
If 4 years is one half-life, then 12 years is 3 half-lives. During the first half-life—the first 4 years—half of the sample will decay. During the second half-life (years 4 to 8), half of the remaining half will decay, leaving one-fourth of the original. During the third and final half-life (years 8 to 12), half of the remaining fourth will decay, leaving one-eighth of the original sample. Thus, the fraction remaining after 3 half-lives is .
Let n be the number of radioactive nuclei that have not yet decayed in a sample. It turns out that the rate at which the nuclei decay, is proportional to the number that remain (n). This suggests the equation
where λ is known as the decay constant. The solution of this equation tells us how the number of radioactive nuclei changes with time. The is known as an exponential decay:
n = n0e−λt
where n0 is the number of undecayed nuclei at time t = 0. The decay constant is related to the half-life by
A typical exponential decay curve is shown in Figure 9.5.
Figure 9.5 Exponential Decay
If at time t = 0, there is a 2 mole sample of radioactive isotopes, how many nuclei remain after 45 minutes, assuming a decay constant of 2 hr−1?
This question is asking for an application of the exponential decay equation:
Raising Euler’s number (e) to an exponent—especially a fractional exponent—is beyond the scope of the math on the MCAT, but the value of is 0.22. Thus, 22% of the original 2 mole sample remains. This constitutes 0.44 mol, which, if multiplied by Avogadro’s number, gives us the number of nuclei remaining:
MCAT Concept Check 9.4:
Before you move on, assess your understanding of the material with these questions.
1. True or False: Nuclear fission and nuclear fusion both release energy.
2. Compare and contrast nuclear fission and nuclear fusion reactions:
Size of Reactant Particles
Change in Nuclear Mass during Reaction (Increase or Decrease)
3. Complete the following chart:
4. How many half-lives are necessary for the complete decay of a radioactive sample?
5. Which type of nuclear decay could be detected in an atomic absorption spectrum?