Energy Levels in Atoms — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Discrete atomic energy levels, photon emission/absorption transitions, the energy difference formula, ionization and binding energy, line spectra analysis, and constant conventions for AP Physics 2 exam questions.

You should already know: Basic atomic structure, photon energy and wave-particle duality, conservation of energy.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Energy Levels in Atoms?

Energy levels in atoms are the discrete, allowed values of internal energy that a bound atom can have, a core result of quantum mechanics: bound electrons can only exist at specific energies, not any arbitrary value. This topic makes up ~3-4% of the total AP Physics 2 exam weight per the official College Board CED, and appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often paired with spectroscopy or photoelectric effect concepts. Standard notation labels the lowest energy (most stable) ground state as $n=1$ with energy $E_1$, which is always negative for bound atoms: we follow the universal convention of setting zero energy equal to a free electron at rest infinitely far from the nucleus. Higher energy excited states are labeled $n=2,n=3,...$ with less negative energies. Synonyms for this topic include discrete electron energy levels and atomic energy quantization. Unlike classical mechanics which predicted electrons could have any energy, experimental evidence from line spectra confirmed quantization, and this topic forms the foundation for all quantum atomic models tested on the AP exam.

2. Energy Transitions and Photon Energy

When an electron moves between two allowed energy levels, energy is strictly conserved. Because energy levels are discrete, the change in the atom's energy $\Delta E$ is exactly equal to the energy of the photon absorbed or emitted during the transition. For any transition from initial energy $E_i$ to final energy $E_f$, the relation is: $$\Delta E_{\text{atom}}=E_f-E_i=\pm E_{\text{photon}}=\pm \frac{hc}{\lambda }$$ The sign follows energy conservation: if the atom emits a photon, it loses energy, so $\Delta E$ is negative ($E_f<E_i$, electron drops to a lower level). If the atom absorbs a photon, it gains energy, so $\Delta E$ is positive ($E_f>E_i$, electron jumps to a higher level). Only photons with energy exactly equal to $|\Delta E|$ can be absorbed or emitted; photons with the wrong energy pass through the atom without interaction. A critical time-saving constant for AP Physics 2 is $hc\approx 1240\text{eVcdotpnm}$, which lets you calculate wavelength directly in nanometers from energy in electron-volts, no unit conversion required.

Worked Example

A hydrogen atom has a ground state energy of $-13.6\text{eV}$ and a second excited state energy of $-1.51\text{eV}$. What is the wavelength of the photon emitted when an electron drops from the second excited state to the ground state?

1. Identify initial and final states: $E_i=-1.51\text{eV}$ (second excited state, initial), $E_f=-13.6\text{eV}$ (ground state, final).
2. Calculate photon energy as the absolute value of the atom's energy change: $E_{\text{photon}}=|E_f-E_i|=|-13.6-(-1.51)|=12.09\text{eV}$.
3. Rearrange the photon energy relation to solve for $\lambda$: $\lambda =\frac{hc}{E_{\text{photon}}}$.
4. Substitute values: $\lambda =\frac{1240\text{eVcdotpnm}}{12.09\text{eV}}\approx 103\text{nm}$.

Exam tip: Always use the 1240 eV·nm value of hc for AP problems, it eliminates unit conversion errors that are common when converting between joules and eV or meters and nanometers.

3. Ionization and Binding Energy

Ionization energy is the minimum energy required to remove an electron from an atom in its ground state, leaving a free electron with approximately zero kinetic energy. By our zero-energy convention, ionization energy equals the absolute value of the ground state energy. Binding energy is the general term for the energy required to remove an electron from any bound energy level (not just the ground state). If an incoming photon has energy greater than the binding energy of the electron, the excess energy becomes kinetic energy of the ejected free electron, a process that links atomic energy levels to the photoelectric effect tested on the AP exam. Mathematically: $$\text{Binding Energy for level }n=|E_n|$$ $$K{E}_{\text{free electron}}=E_{\text{photon}}-|E_n|$$ This relation only holds if $E_{\text{photon}}>|E_n|$; if the photon energy is too low, ionization does not occur.

Worked Example

A lithium ion has an electron in the $n=2$ energy level with energy $-12.0\text{eV}$. A photon of $18.5\text{eV}$ is absorbed by the ion, ejecting the electron. What is the kinetic energy of the ejected electron?

1. Identify the binding energy of the electron in its initial level: $\text{Binding Energy}=|E_2|=|-12.0\text{eV}|=12.0\text{eV}$.
2. Apply conservation of energy: the photon's energy is split between the energy needed to free the electron and kinetic energy of the free electron.
3. Write and solve the energy balance: $KE=E_{\text{photon}}-\text{Binding Energy}=18.5\text{eV}-12.0\text{eV}=6.5\text{eV}$.
4. Confirm physical consistency: the photon energy is greater than the binding energy, so ejection is possible, and kinetic energy is positive, as expected.

Exam tip: If a question asks for ionization energy from an excited level, never default to the ground state ionization energy. Always use the absolute value of the energy of the level the electron starts in.

4. Atomic Line Spectra

The discrete nature of atomic energy levels produces discrete line spectra, rather than the continuous spectra produced by hot blackbodies. There are two common types of line spectra: emission spectra (bright colored lines on a dark background) produced when excited atoms emit photons of specific energies, and absorption spectra (dark lines on a continuous bright background) produced when cool atoms absorb specific photons from a passing continuous light source. Every element has a unique set of energy levels, so it produces a unique spectral "fingerprint" that can be used to identify elements in unknown samples or distant astronomical objects. If electrons are excited up to energy level $n$, the number of unique emission lines (each corresponding to one unique transition between two levels) is given by the combination formula for choosing 2 levels out of $n$: $$N=\frac{n(n-1)}{2}$$ Wavelength is inversely proportional to photon energy, so the longest wavelength photon always comes from the smallest energy difference between any two levels, and the shortest wavelength comes from the largest energy difference.

Worked Example

A gas of atoms has all electrons excited to the $n=4$ energy level. How many unique emission lines can this gas produce as electrons return to the ground state? Which transition produces the shortest wavelength photon, if energy levels are $E_1=-12\text{eV},E_2=-6\text{eV},E_3=-3\text{eV},E_4=-1.5\text{eV}$?

1. We have $n=4$ energy levels accessible for transitions, so substitute into the line count formula: $N=\frac{4(4-1)}{2}=6$ unique emission lines.
2. Shortest wavelength corresponds to the largest photon energy, which comes from the largest energy difference between two levels.
3. The largest energy difference is between the highest excited level ($n=4$) and the ground state ($n=1$): $\Delta E=|E_1-E_4|=|-12-(-1.5)|=10.5\text{eV}$, which is larger than any other transition energy.
4. The transition that produces the shortest wavelength is therefore $4\to 1$.

Exam tip: Remember the inverse relationship between photon energy and wavelength: smaller energy = longer wavelength, larger energy = shorter wavelength. This is a common point of confusion in line spectra questions.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Calculating wavelength from a negative $\Delta E$ for an emission transition, resulting in a negative wavelength. Why: Students mix up the sign convention for the atom's energy change, forgetting photon energy is always positive. Correct move: Always take the absolute value of $\Delta E$ for photon energy, so $E_{\text{photon}}=|E_f-E_i|$ regardless of transition direction.
• Wrong move: Absorbing a 5 eV photon when the energy difference between two levels is 6 eV, then calculating the electron's new energy as $E_i+5\text{eV}$. Why: Students forget that only photons with energy exactly matching the transition energy can be absorbed. Correct move: If the photon energy does not match any allowed transition from the initial level, the photon is not absorbed, and the atom stays in its original state.
• Wrong move: Calculating ionization energy from an excited $n=2$ state as 13.6 eV (the ground state ionization energy of hydrogen). Why: Students confuse the general definition of ground-state ionization energy with ionization from a specific excited level. Correct move: Always use the absolute value of the initial level's energy when asked for ionization energy from that level.
• Wrong move: Calculating the number of spectral lines as $n$ when electrons are excited to level $n$. Why: Students incorrectly memorize the level count as the line count, mixing up levels and transitions. Correct move: If electrons are excited to level $n$, use the combination formula $N=\frac{n(n-1)}{2}$ to get the number of unique transitions (and lines).
• Wrong move: Getting an order of magnitude error in wavelength after converting eV to joules and nanometers to meters. Why: Students forget the hc shortcut and make arithmetic errors during unit conversion. Correct move: Use $hc=1240\text{eVcdotpnm}$ to get wavelength directly in nanometers when energy is in eV, no conversion needed.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

An atom has the following discrete energy levels: $E_1=-10\text{eV}$ (ground state), $E_2=-4\text{eV}$, $E_3=-3\text{eV}$, $E_4=-1\text{eV}$. If the atom is initially in the ground state, which of the following photon energies can the atom absorb? A) 1 eV B) 3 eV C) 7 eV D) 8 eV

Worked Solution: For absorption from the ground state, the photon energy must exactly equal the difference between the ground state and any higher energy level. We calculate all allowed transition energies from the ground state: $E_2-E_1=6\text{eV}$, $E_3-E_1=7\text{eV}$, $E_4-E_1=9\text{eV}$. The only option that matches an allowed transition energy is 7 eV. The other options do not match any allowed transition from the ground state, so the corresponding photons will not be absorbed. The correct answer is C.

Question 2 (Free Response)

A new element has the following energy levels (in eV): $E_1=-12.0$, $E_2=-6.0$, $E_3=-3.0$, $E_4=-1.5$. (a) What is the ground state ionization energy of the element, in eV? (b) A photon of 13.5 eV is absorbed by an atom in the ground state. What is the kinetic energy of the ejected electron, in eV? (c) A collection of these atoms has all electrons excited to the $n=4$ level. How many unique emission lines are produced? Name the transition that produces the longest wavelength photon, and explain your reasoning.

Worked Solution: (a) Ground state ionization energy equals the absolute value of the ground state energy, per our zero-energy convention: $$E_{\text{ion}}=|E_1|=|-12.0\text{eV}|=12.0\text{eV}$$

(b) Conservation of energy gives the kinetic energy of the free electron as the incoming photon energy minus the binding energy of the ground state: $$KE=E_{\text{photon}}-|E_1|=13.5\text{eV}-12.0\text{eV}=1.5\text{eV}$$ The photon has enough energy to ionize the atom, so the excess energy becomes kinetic energy of the ejected electron.

(c) The number of unique emission lines for electrons excited to $n=4$ is: $$N=\frac{n(n-1)}{2}=\frac{4(3)}{2}=6\text{unique lines}$$ Wavelength is inversely proportional to photon energy ($\lambda =hc/E$), so the longest wavelength corresponds to the smallest photon energy. The smallest energy difference between any two levels is the $4\to 3$ transition: $\Delta E=|-3.0-(-1.5)|=1.5\text{eV}$, which is smaller than any other transition energy. Thus, the $4\to 3$ transition produces the longest wavelength photon.

Question 3 (Application / Real-World Style)

Astronomers observe a dark absorption line at 656 nm in the continuous spectrum of a star behind a cool gas cloud. Hydrogen energy levels are given by $E_n=-13.6/n^2\text{eV}$. What energy level transition produces this absorption line, and how does this confirm the cloud contains hydrogen?

Worked Solution: First, calculate the photon energy of the absorption line using $hc=1240\text{eVcdotpnm}$: $$E_{\text{photon}}=\frac{1240\text{eVcdotpnm}}{656\text{nm}}\approx 1.89\text{eV}$$ Cool hydrogen clouds have most electrons in low energy levels; visible absorption lines for hydrogen start from $n=2$. Calculate $E_2=-13.6/(2^2)=-3.4\text{eV}$. The final energy level after absorption is $E_f=E_i+E_{\text{photon}}=-3.4+1.89=-1.51\text{eV}$. Solve for $n$: $n=\sqrt{13.6/1.51}\approx 3$, so this is the $n=2\to n=3$ transition in hydrogen. Hydrogen has a unique set of energy levels, so this absorption line matches the unique spectral fingerprint of hydrogen, confirming the cloud contains hydrogen.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational quantum model of the atom, and you will immediately apply its core rules of discrete energy and energy conservation to the photoelectric effect and nuclear energy levels next in Unit 7. Without mastering the energy transition rules and quantization convention here, you will not be able to correctly solve problems of photon-matter interaction, radioactive decay, or nuclear binding energy, all of which are heavily tested on the AP Physics 2 exam. This topic also connects to wave properties of matter, since discrete energy levels arise from the standing wave nature of bound electrons, a core quantum concept. It also underpins all spectroscopic techniques used in applied fields from astronomy to medical diagnostics.

• Photoelectric Effect
• Nuclear Binding Energy
• Radioactive Decay and Half-Life
• Wave-Particle Duality