Atomic and Nuclear Physics for the MCAT: Everything You Need to Know

Learn essential MCAT topics on atomic and nuclear physics, work through practice problems, and review detailed answers to boost your exam readiness.

(Note: This guide is part of our MCAT Physics series.)

Table of Contents

Part 1: Introduction to atomic and nuclear physics

Part 2: Atomic structure

a) Nuclear structure

b) Components of the nucleus

Part 3: Radioactive Decay

a) Alpha decay

b) Beta decay

c) Gamma decay

d) Half-Life and Exponential Decay

Part 4: High-Yield Terms and Equations

Part 5: Passage-Based Questions and Answers

Part 6: Standalone Questions and Answers

----

Part 1: Introduction to atomic and nuclear physics

Atomic and nuclear physics is a wide-ranging topic that covers the structure and behavior of the individual atom. In this guide, we'll focus on the most important experimental results and the equations that came to describe some of those results.

On the MCAT, atomic and nuclear physics is a medium-yield topic. Getting these concepts down will help you ace any related questions on the test and might even provide some intuition on chemistry and molecular biology topics, too.

----

Part 2: Atomic structure

An atom is the smallest unit of matter that can comprise a chemical element. For a long time, atoms were thought to be the absolute smallest possible units of matter, until it was discovered that they could be divided into constituent charged particles: namely, protons, neutrons, and electrons, which have positive, neutral, and negative charges, respectively. The magnitude of a single proton and a single electron’s charges is 1.6 x 10^-19 C. Since the charge of a single proton is equal to the charge of a single electron, equal numbers of them create a net charge of zero. 

a) Nuclear structure

Atoms consist of a central cluster of protons and neutrons surrounded by electrons. The central cluster of protons and neutrons is called the nucleus. Early models of the atom, such as J.J. Thompson’s “plum pudding” model, did not include the nucleus and instead proposed that protons, neutrons, and electrons were evenly distributed throughout the atom. Under the plum pudding model, these three types of particles would be homogeneously mixed about and appear to be present at the same density.

Rutherford gold foil experiment

The Rutherford gold foil experiment showed that a dense, positively charged clump of mass must be located at the center of an atom. 

The experimental setup consisted of a very thin sheet of gold foil and a device that would shoot positively charged alpha particles at the foil. (More on alpha particles later, but suffice to say that they are positively charged particles.) If the plum pudding model were correct, the alpha particles would have passed through the foil with no deflection. However, to the researchers’ surprise, a small proportion of the alpha particles experienced a very strong deflection. This suggested the existence of a positively charged nucleus.

Figure 1    The plum pudding model's prediction as compared to the results of the gold foil experiment

The Bohr Model

By the time the results of the gold foil experiment were known, J.J. Thompson had already discovered the existence of negatively charged electrons. These electrons appeared to exist on the periphery of the atom, but no one had definitively proven their location around the nucleus. This inference was eventually developed into the Bohr Model of the atom, which proposed that electrons orbited the nucleus. This model was analogous to the orbits of planets in our Solar System, where the nucleus was substituted for the sun, electrons for the planets, and electrostatic attraction for gravitational attraction. 

Figure 2    In the Bohr model, electrons orbit the nucleus.

Just like planetary orbits, the energy of an electron affects its dynamics. In planetary orbit, a “lower energy” orbit refers to an orbit with a smaller radius that feels a stronger gravitational pull, while a “higher energy” orbit refers to an orbit with a larger radius that feels a weaker gravitational pull. A lower energy orbit is more stable, and a higher energy orbit is less stable. The same principles apply to electron orbit under the Bohr model. 

A key difference from planetary orbits is that the Bohr model states that the possible energy levels are quantized, meaning they are not continuous and can only be certain numbers. The energy levels of the orbiting electrons in a hydrogen atom must follow:

E = −13.6 eV/n²
where E = energy,
n = energy level of the electron

Note that -13.6 eV (electron volts) is the energy of an electron in the ground state of a hydrogen atom. (In general, this energy depends on the square of the atomic number and the Rydberg constant.)

This energy can also be thought of as the energy stored in the bond between electron and nucleus. The more negative the energy, the more stable the state. Notice that higher n makes the energy less negative and less stable. When E becomes zero, the electron dissociates. For other elements, a more general equation replaces the numerator with the Rydberg constant (R) and the atom’s atomic number squared (Z²).

Remember that energy must always be conserved. So, in order for the electron to change energy levels, it needs to either absorb energy or emit energy. The Bohr model specifies that emitted energy is in the form of electromagnetic radiation: most commonly, emitted as visible light. In general, the absorbed energy tends to be electromagnetic radiation, too. 

The type of light absorbed or emitted by the electron is related to the change in energy level because the energy contained in light depends on its frequency. Higher frequency light (or colors that are closer to the color blue in the visible spectrum) has more energy than lower frequency light (like the color red in the visible spectrum). The energy of different frequencies of light is given by:

E = hν = hc/λ
where E = energy,
h = Planck's constant, equal to 6.6 × 10⁻³⁴ m² kg/s,
ν = frequency of light,
c = speed of light in a vacuum, equal to 3 × 10⁸ m/s,
λ = wavelength of the light

Thus, when an electron absorbs blue light, it will jump up more energy levels than an electron that absorbs red light. The relationship between the wavelength of absorbed or emitted light and the change in energy level is given directly by the Rydberg formula:

1/λ = R_H(1/n_f² − 1/n_i²)
where λ = wavelength of the light,
R_H = the Rydberg constant, equal to 1.09 × 10⁷ m⁻¹,
n_f = final energy level,
n_i = initial energy level

Figure 3     As an electron moves between levels, it either gains or loses energy

For most chemists, the Bohr model eventually became outdated and was replaced by the valence shell theory. However, for most systems—the hydrogen atom in particular—it is still a good enough approximation of the complex quantum mechanics that are involved. 

b) Components of the nucleus

The nucleus of an atom holds the protons and neutrons and defines the atom’s most important characteristics. The atomic number describes how many protons are in the nucleus. It defines what element the atom is: two atoms with the same atomic number must be the same element. 

The total atomic mass is the number of protons and neutrons in the nucleus. It’s worth noting that protons and neutrons both have the same mass: approximately 1 amu, or atomic mass unit. 

Unlike the number of protons, the number of neutrons in a single element may vary. These variations are called isotopes. Sometimes isotopes are specified by writing the atom as the elemental symbol, dash, then a number. That number refers to the atomic mass of that isotope (O¹⁶ has 8 protons and 8 neutrons, for example). The atomic weight may also be used for elements that have multiple stable isotopes. It is the average of the atomic masses of each stable isotope—so it might be a decimal! 

Figure 4     How to write isotopic notation

Nucleons (a catch-all name for protons and neutrons) are held together very tightly by the strong nuclear force. The energy required to split them all apart is called the binding energy of an atom. This energy is huge. Einstein’s famous equation E=mc² applies here—note that a very small amount of mass m is required to create a huge amount of binding energy E, since m is multiplied by the speed of light c squared.

Perhaps surprisingly, there is a difference between the experimental mass of an atom that is measured and the predicted mass that results by adding individual masses of nucleons together. The leftover mass in this equation accounts for the mass defect. The experimental mass is always less than predicted because some mass is absorbed by the immense energy of the bonds between nucleons.

In other words,

m_D = m_P − m_e and m_D = E/c²

The mass defect, and therefore the binding energy, is different for different elements. The binding energy per nucleon is highest for iron, with atomic number 56.

In addition to the strong nuclear force, there is a weak nuclear force that serves to hold subatomic particles together. However, as the name implies, it is much weaker than the strong nuclear force and plays a much smaller role. 

----

Part 3: Radioactive Decay

So far, we’ve mentioned the “stability” of atoms a few times. We’ve described the electromagnetic emission that occurs when electrons become too unstable, but what happens when the nucleus is too unstable? 

This is when radioactive decay occurs. Radioactive decay occurs when the nucleus splits off particles or energy in order to stabilize itself. There are three types to know on the MCAT. For all types, mass and charge are conserved during the decay. Thus, when you see reaction equations, the atomic numbers and atomic masses on each side must add up to the same number.

a) Alpha decay

The first type of decay is alpha decay, in which the original unstable atom emits an alpha particle at a low speed. An alpha particle is a cluster of two protons and two neutrons—essentially the nucleus of a helium atom. The reaction equation is below:

^A_ZX → ^A−4_Z−2Y + ⁴₂α
where X = atomic number of original element,
Z = initial atomic number,
Y = resulting element,
α = alpha particle, equivalent to a helium nucleus

Alpha decay is the lowest energy radioactive decay. The emitted particle moves relatively slowly and can be stopped by something as thin as a piece of paper.

b) Beta decay

The next type of decay is beta decay. During beta decay, the nucleus shoots off a beta particle at a high speed. A beta particle is a fast-moving electron or positron (a particle of low mass with a positive charge). As a result, there are two subtypes of beta decay: positron emission and electron emission. These are also referred to as 𝛃⁺ and 𝛃⁻ decay.

You might be thinking, “Where in the nucleus does the electron come from?” It’s true that there are no standalone electrons in the nucleus, but when this decay occurs, it turns a neutron into a proton, an electron, and an antineutrino (this is just for thoroughness—you don’t have to worry about the antineutrino), and ejects the electron but keeps the proton. For this reason, 𝛃⁻ decay increases the atomic number by one but keeps the atomic mass the same.

^A_ZX → ^A_Z+1Y + ⁰₋₁β

The opposite is true for β⁺ decay:

^A_ZX → ^A_Z−1Y + ⁰₊₁β

Beta decay is a higher energy radiation than alpha decay; beta particles are able to pass through paper but not aluminum foil.

c) Gamma decay

The last type of decay is gamma decay, the highest energy radioactive decay. Gamma decay is unique because no particles are ejected from the nucleus. Instead, high-energy gamma photons are ejected. The composition of the atom does not change; rather, the atom enters a more relaxed, low-energy state. 

Gamma decay is the highest energy radioactive decay, and you would need a lead barrier to stop it. As a result, it is also considered the most destructive. Gamma decay can also destroy DNA and lead to irreversible genetic damage. 

d) Half-life and exponential decay

A sample’s rate of decay depends on two things. The first is the stability of that element’s nucleus. Some elements are inherently unstable. For example, all elements with more than 82 protons (i.e., anything “heavier” than lead) will eventually decay. Some normally stable elements may have unstable isotopes, such as oxygen’s O-19 isotope. The instability of a certain element or isotope is quantified through its half-life. 

Half-life is the amount of time it takes for half of the original sample to decay. After the length of two half-lives, one-half of the remaining portion decays, leaving one-quarter of the original sample, and so on. Very unstable elements will have very small half-lives, and less unstable elements will have longer half-lives. (Elements that are actually stable do not have half-lives because they do not undergo radioactive decay.)

The MCAT will often test you on questions about half-life and the amount of time it takes for a sample to decay. For these questions, creating a table of half-lives may help you solve for the amount of a sample that is remaining or the amount of time that must have elapsed. 

Figure 5    Half life table

The second factor that contributes to the rate of decay is how much of that sample is left. The rate of decay of the sample is directly proportional to the amount of sample that is left. This relationship describes exponential decay, one that starts off dramatically—with many quick emissions—and becomes less active over time. The precise shape of an atom’s exponential decay depends on both the initial amount and the half-life. They come together in the equation:

N = N₀ × (1/2)^{# of half-lives} = N₀ × (1/2)^t/τ
where N = the amount of sample left at time t,
N₀ = initial amount of sample,
t = time after start,
τ = length of half-life

Figure 6     Graphical display of halflife table

You might also see exponential decay written in terms of a decay constant. This other form is as seen below:

N = N₀ × e^λt
where N = the amount of sample left at time t,
N₀ = initial amount of sample,
e = Euler's number, approximately 2.72,
t = time after start,
λ = the decay constant

The decay constant can be related to the half-life by the following formula:

τ = ln(2)/λ
where τ = length of half-life,
λ = the decay constant,
ln(2) = the natural logarithm of 2

While you won’t need to memorize each of these equations, it’s important to be familiar with these relationships and the shape of the exponential decay graph. These trends are often plotted on a semilog plot, in which one axis uses a logarithmic scale and the other uses a linear scale. In the context of radioactive decay, the y-axis usually depicts radioactive counts (e.g., in Curies) and uses a logarithmic scale. The x-axis usually depicts time (e.g., in years or thousands of years) and uses a linear scale. 

The semilog plot is used in contrast to a log-log plot, which uses a logarithmic scale on both the x- and y-axes. So, it is important to pay attention to the units used on each graph you may come across!

----

Part 4: High-yield terms and equations

Atom: the smallest unit of matter that can comprise a chemical element

Proton: a positively charged particle in the nucleus

Neutron: a neutrally charged particle in the nucleus

Electron: a negatively charged particle that orbits the nucleus

Nucleon: a particle in the nucleus (a proton or a neutron)

Plum pudding model: an early model of the atom in which all constituent particles were distributed throughout the atom

Gold foil experiment: proved the existence of a positively charged nucleus, disproving the plum pudding model

Bohr model: a model of the atom in which electrons orbit the nucleus in elliptical orbits; outdated but still provides reasonable approximation for the hydrogen atom

Energy quantization: when there are discrete, not continuous, possible energy levels

Dissociation: when an electron gains too much energy and escapes the atom

Ionization energy: the energy required to make the highest energy electron dissociate 

Atomic number: the number of protons in an atom; defines the element

Atomic mass: the number of nucleons in an atom

Isotope: variations of the same element that have different atomic masses

Atomic weight: the average atomic mass of all the natural isotopes of an element

strong nuclear force: the force that binds protons and neutrons in the nucleus

Binding energy: the energy required to separate the nucleons of an atom

Mass defect: the difference in predicted and experimental mass due to e=mc^2

Experimental mass: the measured mass of an atom

Predicted mass: the mass calculated from adding the individual weights of nucleons

Fission: a process that splits a nucleus into two smaller nuclei

Fusion: a process that fuses multiple nuclei into a single nucleus

Alpha decay: radioactive decay in which a nucleus emits an alpha particle at a low speed

Alpha particle: a particle made of two protons and two neutrons, basically a helium nucleus

Beta decay: radioactive decay in which a nucleus emits a beta particle at a high speed

Beta particle: a fast-moving electron

Gamma decay: radioactive decay in which a nucleus emits a gamma photon

Exponential decay: a decay that starts off fast but slows with time

Half-life: the time it takes for half the original sample to decay

Decay constant: a constant that allows you to write the half-life equation as a power of e

Equations

Electron energy levels: E = −13.6 eV/n²

Absorbed or emitted light: 1/λ = R_H(1/n_f² − 1/n_i²)

Absorbed or emitted light: E = hν = hc/λ

Mass defect: m_D = m_P − m_e and m_D = E/c²

Alpha decay: ^A_ZX → ^A−4_Z−2Y + ⁴₂α

Beta-minus decay: ^A_ZX → ^A_Z+1Y + ⁰₋₁β

Beta-plus decay: ^A_ZX → ^A_Z−1Y + ⁰₊₁β

Rydberg constant: R_H = 1.09 × 10⁷ m⁻¹

Planck's constant: h = 6.63 × 10⁻³⁴ Js

Speed of light in a vacuum: c = 3 × 10⁸ m/s

----

Part 5: Passage-based questions and answers

Nuclear power plants account for twenty percent of total energy production in the United States. Most of these power plants use uranium-235 for fuel. When bombarded with neutrons, uranium-235 may absorb one neutron, briefly becoming uranium-236, before undergoing a fission that produces krypton-89, barium-144, and several neutrons. Each individual fission produces energy on the order of 3 x 10^-11 J. One kilogram of uranium-235 can produce around 2-3 million times the energy of a kilogram of coal.

In order to sustain a reaction of uranium-235 fission, however, the proportion of uranium-235 to other uranium isotopes in the fuel must be very high. Uranium-235 makes up less than one percent of natural uranium deposits. Uranium-238 and uranium-234 are other naturally occurring isotopes, with the vast majority of natural deposits composed of uranium-238. Uranium-238 and uranium-234 have half-lives of 4.6 billion years and 245,000 years, respectively. To collect usable fuel, each must be removed from uranium deposits through an enrichment process. The excess uranium-238 and uranium-234 are then disposed of in waste sites.

Because uranium-238 and uranium-234 are still radioactive and undergo alpha decay, the farmers that are located close to nuclear power plants complain that the resulting radioactive activity may contaminate and decrease the quality of their produce. To analyze the effects on their production, three farmers plot three equally sized farms that are each one mile away from a waste site. The production of each farm in the first three years after the establishment of the waste sites and the contents of each waste site are listed in Figure 1.

Figure 1

This passage was produced for practice purposes only

Question 1: Which of the following equations correctly depicts the described lowest energy decay reaction for uranium-238?

A) ⁹²₂₃₈U → ⁹²₂₃₄U + ²₄ɑ
B) ⁹²₂₃₈U → ⁹⁰₂₃₄Th + ²₄ɑ
C) ⁹²₂₃₈U → ⁹²₂₃₉U + ^-1₀β
D) ⁹²₂₃₈U → ⁹³₂₃₈Np + ^-1₀β

Question 2: A researcher suggests that the farms are contaminated. Based on the information in the passage, a reasonable conclusion would be:

A) Abernathy farm is affected because its production decreases from 2000 to 2001

B) No farms are affected because the waste is not radioactive

C) No farms are affected because the half-life of the waste is so large

D) Johnson Farm is affected because it is near the less stable uranium-234

Question 3: If these farms and these waste sites were observed indefinitely, which farm is most likely to be affected first?

A) Abernathy Farm

B) Johnson Farm

C) Williams Farm

D) They would all be affected at around the same time

Question 4: Suppose the production of radioactive waste at the site near Johnson Farm stops. How much of its original contents can be expected to remain in approximately 1 million years?

A) 50%

B) 75%

C) 25%

D) 6%

Answer key for passage-based questions

1. Answer choice B is correct. The passage states that U-238 undergoes alpha decay. The correct decay will have U-238 splitting into an alpha particle (which has atomic number 2 and atomic mass 4), and an element whose atomic number is 2 fewer than U-238 and whose atomic mass is 4 lower than U-238. Th-234 is the only possible choice (choice B is correct.) Uranium would not retain 92 protons after alpha decay (choice A is incorrect). Radioactive decay obeys the conservation of mass, and each side of a decay expression must be balanced (choice C is incorrect). Beta decay of uranium would result in neptunium (choice D is incorrect). 

2. Answer choice C is correct. The passage tells us that the half-lives for uranium-238 and 234 are 4.5 billion years and 245,000 years, respectively. So, in just a few years, there is no significant probability of enough uranium decaying to cause any real effect (choice C is correct). Although Abernathy farm does lose some production after the first year, this correlation does not imply causation, and production does appear to rebound in the next year (choice A is incorrect). All uranium isotopes are, in fact, radioactive (choice B is incorrect). Uranium-234 is less stable than uranium-238, and Johnson farm does show a trend, but the half-life of U-234 is too long to be implicated in that downward trend (choice D is incorrect). 

3. Answer choice B is correct. If these sites were observed indefinitely, any farm located near significant quantities of uranium-234 would likely be affected first. This is because the half-life of U-234 is drastically lower than the half-life of U-238 (choice B is correct). 

4. Answer choice D is correct. To find the answer, it may be helpful to construct a table. After one half-life (or 245,000 years), approximately 50% of the original contents should remain (choice A is incorrect). After two half-lives (or about half a million years), approximately 25% of the original contents should remain (choice C is incorrect). After three half-lives (or about ¾ of a million years), approximately 12.5% of the original contents should remain. After four half-lives (or about one million years), approximately 6% of the original contents should remain (choice D is correct). 

----

Part 6: Standalone questions and answers

Question 1: Which of the following best accounts for the emission of discrete emission lines by a hydrogen atom leaving an excited state?

A) Strong nuclear force

B) Mass defect

C) Nuclear fission

D) Energy quantization

Question 2: Rutherford’s gold foil experiment discovered repulsion between two particles that are due to which of the following forces?

A) Strong nuclear force

B) Weak nuclear force

C) Electromagnetic force

D) Gravitational force

Question 3: Recall that the energy level of an electron in the ground state of hydrogen is -13.6 eV. If an electron then moves to energy level 2, how much more energy must be absorbed before the electron dissociates?

A) 13.6 eV

B) 6.8 eV

C) 3.4 eV

D) 1.7 eV

Question 4: The mass defect of O-16 is 0.133 amu, and the mass defect of C-12 is 0.099 amu. Which has a higher average binding energy per nucleon?

A) O-16

B) C-12

C) All non-metals have the same average binding energy per nucleon

D) All elements have the same average binding energy per nucleon

Question 5: Which of the following reactions correctly depicts beta decay?‍ ‍

A) ²³⁸₉₂U → ²³⁸₉₁Np + ⁰_-1β
B) ¹³⁷₅₅Cs → ¹³⁷₅₆Ba + ⁰_-1β
C) ¹⁴₆C → ¹³₇N + ¹_-1β
D) ²²⁸₈₈Ra → ²²⁹₈₉Ac + ¹_-1β

Answers to standalone questions

1. Answer choice D is correct. Recall that electrons in an atom can have only certain energy levels. This is referred to as the quantization of energy. When an electron jumps down energy levels, it emits a wavelength of light proportional to this change. There are a finite number of energy levels, so only discrete emission lines can be produced (choice D is correct). The strong nuclear force is the force that binds nucleons (choice A is incorrect). Mass defect refers to the mass lost to the high-energy bonds between nucleons (choice B is incorrect). Nuclear fission occurs when the nucleus splits into two smaller particles (choice C is incorrect).  

2. Answer choice C is correct. Recall that the gold foil experiment led to the discovery of the nucleus. This discovery was made because positively charged alpha particles were deflected by a small, positively charged mass in the center of the atom. Like charges repel because of electromagnetic forces (choice C is correct). The strong nuclear force is the force that binds nucleons (choice A is incorrect). The weak nuclear force is not relevant in this setting (choice B is incorrect). Gravitational force, or gravity, is present between large masses and is not dependent on charge (choice D is incorrect). 

3. Answer choice C is correct. The energy of an electron in the hydrogen atom is given by the equation E=-13.6/n², where n is the energy level. Notice that this energy is negative; when it reaches zero, the electron dissociates. We are told our electron is in energy level 2, so its energy is E=-13.6/(2^2)=-3.4 eV. In order to reach E=0 and dissociate, the electron must absorb 3.4 eV (choice C is correct).

4. Answer choice A is correct. Recall that E=mc², where m is the mass defect and E is the binding energy. We know the mass defect for both, and so can calculate the energy due to binding by multiplying the mass defect m by c². The binding energies for O-16 and C-12 are 0.133*c² and 0.099*c², respectively; to obtain the binding energy per nucleon, we must divide by the number of nucleons, 16 and 12, respectively. 0.133*c²/16=.00831*c² and 0.099c²/12=.00825*c². So, O-16 has the higher binding energy per nucleon (choice A is correct).

5. Answer choice B is correct. Recall that a beta particle has an atomic number of -1 and atomic mass of 0 (choices C and D are incorrect). Thus, the atomic number of the resulting isotope should increase by one, while atomic mass remains the same (choice B is correct).