AP Physics 2 · Topic 15.2

Topic 15.2: The Bohr Model of Atomic Structure

Unit 15: Modern Physics12-15% of the multiple-choice section

The Bohr model treats an electron as orbiting the nucleus in a circle, held there by the electric force acting as the centripetal force. Only some orbits are allowed: the circumference has to be a whole number of the electron's de Broglie wavelengths, which is what makes its energies discrete.

AP Physics: Unit 15 (topics 15.2 The Bohr Model of Atomic Structure). AP Physics 2 Unit 15, Topic 15.2. One learning objective, 15.2.A, describe the properties of an atom. Essential knowledge: 15.2.A.1 (atoms have internal structure) with 15.2.A.1.i (small positively charged nucleus surrounded by one or more negatively charged electrons), 15.2.A.1.ii (nucleus made of protons and neutrons), 15.2.A.1.iii (nuclear notation), 15.2.A.1.iv (an ion is an atom with a nonzero net electric charge); 15.2.A.2 (each atomic element has a unique number of protons) with 15.2.A.2.i (number and arrangements of electrons affects how atoms interact), 15.2.A.2.ii (total number of neutrons and protons identifies the isotope), 15.2.A.2.iii (the mass of an atom is dominated by the total mass of the protons and neutrons in its nucleus); and 15.2.A.3 (the Bohr model of the atom is based on classical physics and was the historical representation of the atom that led to the description of the hydrogen atom in terms of discrete energy states) with 15.2.A.3.i (electrons modeled as moving around the nucleus in circular orbits determined by the electron's charge and mass and the electric force between electron and nucleus; relevant equations F_e = k q1 q2 / r^2 and F_net = m v^2 / r) and 15.2.A.3.ii (the standing wave model of electrons accounts for the existence of specific allowed energy states, because the electron orbit's circumference must be an integer multiple of the electron's de Broglie wavelength). BOUNDARY STATEMENT, printed under this topic: the analysis and description of electron structure is limited to energy levels and will not include such advanced descriptions as orbitals, orbital shapes, or probability functions. Suggested skills: 1.A, 2.B, 2.C, 3.B. The Modern Physics group of the AP Physics 2 equation sheet contains no equation for this topic; the two relevant equations are borrowed from the Electricity group and the Mechanics and Fluids group. No hydrogen energy-level formula, Rydberg formula, Rydberg constant or Bohr radius appears in the CED's required course content or on the equation sheet; E_n = (-13.6 eV)/n^2 appears only in an optional sample instructional activity listed under Topic 15.3.

What Topic 15.2 requires

Topic 15.2 carries a single learning objective, 15.2.A: describe the properties of an atom. The essential knowledge sits in three numbered groups, and the topic does have a boundary statement, printed under the last of them.

15.2.A.1 Atoms have internal structure.
15.2.A.1.i Atoms consist of a small, positively charged nucleus surrounded by one or more negatively charged electrons.
15.2.A.1.ii The nucleus of an atom is made up of protons and neutrons.
15.2.A.1.iii The number of neutrons and protons in an atom can be represented using nuclear notation.
15.2.A.1.iv An ion is an atom with a nonzero net electric charge.
15.2.A.2 Each atomic element has a unique number of protons.
15.2.A.2.i The number and arrangements of electrons affects how atoms interact.
15.2.A.2.ii The total number of neutrons and protons identifies the isotope of an element.
15.2.A.2.iii The mass of an atom is dominated by the total mass of the protons and neutrons in its nucleus.
15.2.A.3 The Bohr model of the atom is based on classical physics and was the historical representation of the atom that led to the description of the hydrogen atom in terms of discrete energy states.
15.2.A.3.i In the Bohr model of the atom, electrons are modeled as moving around the nucleus in circular orbits determined by the electron's charge and mass, as well as the electric force between the electron and the nucleus. Relevant equations: $F_e = k \frac{q_1 q_2}{r^2}$ and $F_{net} = m \frac{v^2}{r}$ .
15.2.A.3.ii The standing wave model of electrons accounts for the existence of specific allowed energy states of an electron in an atom, because the electron orbit's circumference must be an integer multiple of the electron's de Broglie wavelength.

Boundary statement: the analysis and description of electron structure is limited to energy levels and will not include such advanced descriptions as orbitals, orbital shapes, or probability functions.

Suggested skills: 1.A (create diagrams, tables, charts, or schematics to represent physical situations), 2.B (calculate or estimate an unknown quantity with units from known quantities), 2.C (compare physical quantities between two or more scenarios or at different times and locations in a single scenario), and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim).

Notice the shape of the topic. Two thirds of it is plain atomic structure that a chemistry course also covers, and the physics only starts at 15.2.A.3. Do not let that lull you: the two sub-statements under 15.2.A.3 carry the whole argument of the unit, and 15.2.A.3.ii in particular is where quantization stops being an assertion and becomes a consequence.

Structure first: nucleus, electrons, ions

15.2.A.1.i gives the picture the rest of the topic works inside: a small, positively charged nucleus surrounded by one or more negatively charged electrons. The word small is doing real work. Everything the Bohr model calculates about orbits treats the nucleus as a point charge sitting at the centre, and that only works because the nucleus occupies a tiny fraction of the atom's volume.

15.2.A.1.ii fills the nucleus with protons and neutrons, collectively nucleons, which is the vocabulary Topic 15.7 and Topic 15.8 run on.

15.2.A.1.iv is a one-line definition and it is precise: an ion is an atom with a nonzero net electric charge. Note what it does not say. It says nothing about how the atom got that way and nothing about which sign. An atom that has lost an electron and an atom that has gained one are both ions.

15.2.A.2.iii gets its own worked check below, because the claim that the mass of an atom is dominated by the protons and neutrons is easy to accept and easy to underestimate. Using the printed masses, a proton or a neutron is about 1800 times an electron's mass, so for a light atom the electrons contribute a few hundredths of a percent of the total.

The force holding this together is the one you already know. The electric force between the positive nucleus and a negative electron is attractive, follows an inverse square law, and is the only force the Bohr model uses. Gravity between an electron and a proton is many orders of magnitude weaker and never appears in this unit.

Nuclear notation, elements and isotopes

15.2.A.1.iii says the number of neutrons and protons in an atom can be represented using nuclear notation. The CED does not spell the notation out in the essential knowledge text, but it is the standard three-part symbol used throughout Topics 15.7 and 15.8:

^{A}_{Z}\text{X}

where $Z$ is the number of protons (the atomic number), $A$ is the total number of nucleons (the mass number), and the number of neutrons is $A - Z$ . The chemical symbol X is redundant information: it is fixed by $Z$ , because of 15.2.A.2, each atomic element has a unique number of protons.

That gives you three definitions worth keeping apart:

Change what	You get	CED statement
A different number of protons	A different element	15.2.A.2
A different number of neutrons, same protons	A different isotope of the same element	15.2.A.2.ii
A different number of electrons, same nucleus	An ion of the same isotope	15.2.A.1.iv

15.2.A.2.i adds the chemistry-facing consequence: the number and arrangements of electrons affects how atoms interact. Two isotopes of an element behave almost identically in a chemical reaction because they carry the same electrons; they differ in mass, and in whether the nucleus is stable, which is Topic 15.8's subject.

One cross-reference worth holding on to now. The boundary statement of Topic 15.3 says that in AP Physics 2, only energy level diagrams of single-electron atoms will be considered. Single-electron does not mean hydrogen only. A doubly ionized lithium atom has three protons and one electron, and it is exactly the sort of system the exam can draw a level diagram for.

The Bohr model is built out of classical physics

15.2.A.3 is unusually candid for a curriculum document: the Bohr model of the atom is based on classical physics and was the historical representation of the atom that led to the description of the hydrogen atom in terms of discrete energy states.

Read that as three separate claims.

It is classical. The orbit itself, the force, the acceleration, the speed: all of it is Newtonian mechanics plus Coulomb's law. Nothing quantum has entered yet.
It is historical. The CED is telling you this is a model of record, not the current description of an atom. That is also why the boundary statement then bars orbitals and probability functions: those belong to the model that replaced it.
It led to discrete energy states. Its value is that it produces the right answer for hydrogen's energy levels, which is what Topic 15.3 then observes as a spectrum.

15.2.A.3.i names the machinery exactly: electrons are modeled as moving around the nucleus in circular orbits determined by the electron's charge and mass, as well as the electric force between the electron and the nucleus. Its two relevant equations are borrowed from elsewhere on the sheet:

F_e = k \frac{q_1 q_2}{r^2} \qquad F_{net} = m \frac{v^2}{r}

The first is Coulomb's law from the Electricity group. The second is the centripetal form of Newton's second law from circular motion in AP Physics 1. Setting them equal is the entire dynamical content of the Bohr model:

k \frac{e^2}{r^2} = \frac{m_e v^2}{r}

for a single electron of charge magnitude $e$ orbiting a single proton. One equation, two unknowns ( $v$ and $r$ ), which is precisely the problem. Classical physics allows any radius, with a matching speed. Nothing in this equation picks out a list.

The Modern Physics group of the equation sheet contains no equation for this topic at all. The two the CED cites live in other groups. That is worth knowing before an exam: if you are hunting for a Bohr equation among the ten modern physics lines, you will not find one.

The standing-wave condition, and where the quantization comes from

15.2.A.3.ii is the sentence that turns a continuum of classical orbits into a list, and it is worth having word for word: the standing wave model of electrons accounts for the existence of specific allowed energy states of an electron in an atom, because the electron orbit's circumference must be an integer multiple of the electron's de Broglie wavelength.

Written out, with $\lambda$ the de Broglie wavelength from Topic 15.1:

2\pi r = n\lambda = n \frac{h}{m_e v} \qquad n = 1, 2, 3, \dots

The analogy the CED is invoking is a standing wave on a loop rather than on a string. Bend a string round into a circle and join the ends. A wave running round it only survives if, after one lap, it arrives back in step with itself. If the circumference is not a whole number of wavelengths, the wave meets its own tail out of phase on every lap and cancels. On a loop there is no fundamental-plus-overtones series set by fixed ends; there is just the closure condition, and it admits exactly the integer cases.

Now count the equations. You have the force balance from 15.2.A.3.i, and you have the closure condition from 15.2.A.3.ii. Two equations, two unknowns, one integer $n$ . The system is solved, and the solutions come in a discrete family indexed by $n$ . That is the mechanism behind 15.1.A.5, which asserted without proof that energy and momentum take discrete values for bound systems.

A short chain worth memorising in this order, because it is the shape of a 3.B answer:

Matter has a wavelength, $\lambda = h/p$ (15.1.A.4.i).
A bound electron's wave has to close on itself round the orbit (15.2.A.3.ii).
Only certain radii let it close, so only certain speeds and energies exist.
So the atom's energies are a list with gaps (15.1.A.5).
So the photons it emits and absorbs are a list too (15.3.A.2).

That is the whole argument, and none of it needs a memorised constant.

Energy levels, and what the CED does not give you

The boundary statement fixes the level of description: the analysis and description of electron structure is limited to energy levels and will not include such advanced descriptions as orbitals, orbital shapes, or probability functions.

That is a genuine reduction in workload. On this exam an atom's electronic structure is a ladder of energies, drawn as horizontal lines. It is not s, p and d orbitals, not electron configurations, not shapes, not probability clouds. If a question shows you an atom's structure, it will show you a level diagram.

What is worth saying just as loudly is what the CED does not supply for those levels.

There is no hydrogen energy-level formula in the required course content. The expression $E_n = \frac{-13.6 \text{ eV}}{n^2}$ appears exactly once in the whole AP Physics 2 CED, and it is not in Topic 15.2. It is inside an optional Sample Instructional Activity for Topic 15.3, in a demonstration where teachers show a hydrogen discharge tube and tell students that the energy levels of hydrogen can be modeled that way. The activities page states that those activities are optional and that teachers do not need to use them.
There is no Rydberg formula and no Rydberg constant, anywhere in the CED or on the equation sheet.
There is no Bohr radius printed on the sheet.
The Modern Physics group of the sheet prints no equation for this topic.

The practical rule that follows: if a question needs numerical energy levels, it will give them to you, as a diagram or as a formula in the stem. Treat $-13.6$ eV the way the Topic 15.5 boundary statement tells you to treat work functions, as data the question supplies rather than a constant you carry in.

What you can always do is build from printed constants. The third worked example below derives an allowed orbit radius using only $h$ , $m_e$ , $k$ and $e$ , all four of which are printed in the Table of Information, together with the CED's own two relevant equations and the closure condition. That derivation is more than Topic 15.2 asks of you, but it demonstrates that the quantization is a result rather than a rule handed down.

Reading and drawing the model (skills 1.A and 2.C)

Skill 1.A, create diagrams, tables, charts, or schematics to represent physical situations, is listed for this topic, and there are only two diagrams it can reasonably mean.

The orbit diagram. A nucleus at the centre, an electron on a circle, and two labelled vectors: the electric force pointing from the electron toward the nucleus, and the velocity tangent to the circle. Get those two perpendicular and get the force pointing inward, and you have said everything 15.2.A.3.i says. A common error is drawing the force along the direction of motion, which would speed the electron up rather than turn it.

The standing wave on the orbit. The same circle with a wave drawn along it, closing smoothly after a whole number of wavelengths. Drawing a case that does not close is often the more useful diagram, because it shows why that radius is excluded.

Skill 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario, is the other listed skill, and it is the one that suits a multiple-choice question. The comparisons the model supports, all reading straight off $k e^2/r^2 = m_e v^2/r$ :

Move the electron to a larger radius	What happens
Electric force on it	Decreases, as $1/r^2$
Orbital speed	Decreases, as $1/\sqrt{r}$ , since $v^2 = k e^2 / (m_e r)$
Momentum	Decreases, so the de Broglie wavelength increases (15.1.A.4.i)
Kinetic energy	Decreases, as $1/r$
Total energy of the bound system	Increases, toward zero, because a more distant electron is less tightly bound

That last row is the one to be careful with, and it is where electric potential energy earns its place. A bound system has negative total energy on the usual convention that the energy is zero when the two charges are infinitely far apart. Moving the electron outward makes the total energy less negative, which is an increase, even though the speed and the kinetic energy both went down. Say "less negative" in an answer if you want to be unambiguous.

How Topic 15.2 is tested, and where it goes next

The AP Physics 2 exam is 3 hours long: 42 multiple-choice questions in 85 minutes, then 4 free-response questions in 95 minutes, each section worth 50 percent. Unit 15 as a whole carries a 12 to 15 percent weighting on the multiple-choice section. A four-function, scientific, or graphing calculator is allowed on both sections.

What the four listed skills imply about the questions:

2.B, calculate. Given a radius, get a force or a speed. Given a nuclear symbol, get the number of neutrons. These are short, and the arithmetic is the whole difficulty: $e^2 = 2.56 \times 10^{-38}$ and small radii squared are where sign-of-exponent errors live.
2.C, compare. Two orbits, two atoms, two isotopes. Which has the greater speed, force, wavelength, mass.
3.B, apply a model to make a claim. Why can the electron not sit at an arbitrary radius. Why does the standing-wave condition produce discrete states. These want the argument, not a number.
1.A, represent. Draw it.

A reasonable checklist before leaving the topic. Can you state what makes two atoms different elements, different isotopes, or one an ion of the other, using the CED's own three statements? Can you set the electric force equal to the centripetal force and solve for the speed? Can you say, in two sentences, why a wave that has to close on itself produces a discrete set of orbits? Can you say what the boundary statement rules out?

Where it goes: Topic 15.3 takes the energy levels this topic justifies and turns them into something observable. The transition between two levels emits or absorbs a photon whose energy equals the difference, and the set of those differences is the element's spectrum. Topic 15.2 is the reason the levels exist; Topic 15.3 is the evidence that they do.

Nuclear notation, isotopes, ions, and where the mass is

A lithium nucleus is written in nuclear notation with mass number $A = 7$ and atomic number $Z = 3$ . (a) How many protons, neutrons and electrons does the neutral atom have? (b) A doubly ionized version of this atom has lost two electrons. State its net charge and say why it is a system Topic 15.3 can draw an energy level diagram for. (c) Using the printed masses, find the fraction of the neutral atom's mass carried by its electrons.

(a) By 15.2.A.2, the atomic number is the proton count, so there are $Z = 3$ protons. By 15.2.A.1.iii the mass number counts all nucleons, so the neutron count is $A - Z = 7 - 3 = 4$ . A neutral atom has as many electrons as protons, so 3 electrons.
(b) Removing two electrons leaves 3 protons and 1 electron, a net charge of $+2e = 2(1.60 \times 10^{-19} \text{ C}) = 3.20 \times 10^{-19} \text{ C}$ . By 15.2.A.1.iv, an atom with a nonzero net electric charge is an ion.
It is a legitimate AP Physics 2 energy-level system because it has exactly one electron left, and the Topic 15.3 boundary statement says only energy level diagrams of single-electron atoms will be considered. Single-electron does not mean hydrogen.
(c) Use the printed masses: $m_p = m_n = 1.67 \times 10^{-27} \text{ kg}$ and $m_e = 9.11 \times 10^{-31} \text{ kg}$ . Total nucleon mass $= 7(1.67 \times 10^{-27}) = 1.169 \times 10^{-26} \text{ kg}$ .
Total electron mass $= 3(9.11 \times 10^{-31}) = 2.73 \times 10^{-30} \text{ kg}$ .
Fraction $= \dfrac{2.73 \times 10^{-30}}{1.169 \times 10^{-26} + 2.73 \times 10^{-30}} = 2.3 \times 10^{-4}$ , about $0.023$ percent. That is 15.2.A.2.iii made numerical: the mass of an atom is dominated by the total mass of the protons and neutrons in its nucleus.
One consequence worth carrying forward: ionizing an atom changes its charge dramatically and its mass hardly at all. Losing two of three electrons here removes about 0.016 percent of the mass.

(a) 3 protons, 4 neutrons, 3 electrons. (b) Net charge $+3.20 \times 10^{-19} \text{ C}$ ; with one electron remaining it is a single-electron atom, which is what the Topic 15.3 boundary statement permits. (c) The electrons carry about $2.3 \times 10^{-4}$ , or 0.023 percent, of the atom's mass.

The electric force as the centripetal force

A single electron moves in a circular orbit of radius $r = 1.0 \times 10^{-10} \text{ m}$ around a single proton. Using the two relevant equations the CED lists in 15.2.A.3.i, find (a) the magnitude of the electric force on the electron and (b) its orbital speed.

Set the sign convention before anything else: the positive direction is radially inward, toward the nucleus, because that is the direction of both the electric force on the electron and its centripetal acceleration. All magnitudes below are positive in that sense.
(a) Coulomb's law with $q_1 = q_2 = e = 1.60 \times 10^{-19} \text{ C}$ and the AP sheet's $k = 9.0 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2$ : $F_e = k\dfrac{e^2}{r^2}$ .
Do the squares separately to keep the exponents honest. $e^2 = (1.60 \times 10^{-19})^2 = 2.56 \times 10^{-38} \text{ C}^2$ , and $r^2 = (1.0 \times 10^{-10})^2 = 1.0 \times 10^{-20} \text{ m}^2$ .
So $F_e = \dfrac{(9.0 \times 10^9)(2.56 \times 10^{-38})}{1.0 \times 10^{-20}} = \dfrac{2.304 \times 10^{-28}}{1.0 \times 10^{-20}} = 2.30 \times 10^{-8} \text{ N}$ , directed inward.
(b) This force is the net force on the electron and it is centripetal, so $F_e = m_e \dfrac{v^2}{r}$ . Rearranged, $v^2 = \dfrac{F_e r}{m_e} = \dfrac{(2.304 \times 10^{-8})(1.0 \times 10^{-10})}{9.11 \times 10^{-31}} = \dfrac{2.304 \times 10^{-18}}{9.11 \times 10^{-31}} = 2.53 \times 10^{12} \text{ m}^2/\text{s}^2$ .
Take the root: $v = 1.59 \times 10^6 \text{ m/s}$ . That is about 0.5 percent of $c$ , so treating the electron non-relativistically is fine.
Now notice what has and has not been determined. Nothing in this calculation objected to the radius. Feed in a different $r$ and you get a different force and a different speed, both perfectly consistent. Classical physics allows a continuum of orbits, which is exactly the problem 15.2.A.3.ii exists to fix.

(a) $F_e = 2.30 \times 10^{-8} \text{ N}$ , directed toward the nucleus. (b) $v = 1.59 \times 10^6 \text{ m/s}$ . Any radius would have worked, which is why the model needs an extra condition.

Testing the standing-wave condition, then solving for an allowed radius

Continue the previous example. (a) Test whether $r = 1.0 \times 10^{-10} \text{ m}$ satisfies 15.2.A.3.ii, that the orbit circumference is an integer multiple of the electron's de Broglie wavelength. (b) Then find the smallest radius that does satisfy it, using only constants printed on the equation sheet.

(a) From the previous result, $p = m_e v = (9.11 \times 10^{-31})(1.59 \times 10^6) = 1.45 \times 10^{-24} \text{ kg} \cdot \text{m/s}$ .
The de Broglie wavelength, from the printed $\lambda = h/p$ : $\lambda = \dfrac{6.63 \times 10^{-34}}{1.45 \times 10^{-24}} = 4.58 \times 10^{-10} \text{ m}$ .
The circumference is $2\pi r = 2\pi(1.0 \times 10^{-10}) = 6.28 \times 10^{-10} \text{ m}$ . The ratio is $\dfrac{6.28 \times 10^{-10}}{4.58 \times 10^{-10}} = 1.37$ , which is not an integer. So this orbit is not allowed: after one lap the electron's wave arrives 0.37 of a wavelength out of step with itself and cancels.
(b) Impose both conditions at once. Force balance gives $v = \sqrt{\dfrac{k e^2}{m_e r}}$ , and the closure condition with $n = 1$ gives $2\pi r = \dfrac{h}{m_e v}$ .
Substitute the first into the second and solve for $r$ . $2\pi r = \dfrac{h}{m_e}\sqrt{\dfrac{m_e r}{k e^2}}$ , so $2\pi\sqrt{r} = \dfrac{h}{\sqrt{m_e k e^2}}$ , and therefore $r = \dfrac{h^2}{4\pi^2 m_e k e^2}$ .
Evaluate with printed values only. Numerator: $h^2 = (6.63 \times 10^{-34})^2 = 4.396 \times 10^{-67}$ . Denominator: $m_e k e^2 = (9.11 \times 10^{-31})(9.0 \times 10^9)(2.56 \times 10^{-38}) = 2.099 \times 10^{-58}$ , and $4\pi^2 = 39.48$ , so the denominator is $(39.48)(2.099 \times 10^{-58}) = 8.29 \times 10^{-57}$ .
Divide: $r = \dfrac{4.396 \times 10^{-67}}{8.29 \times 10^{-57}} = 5.30 \times 10^{-11} \text{ m}$ . Because $r$ carries a factor of $n^2$ , the next two allowed radii are $4r = 2.12 \times 10^{-10} \text{ m}$ and $9r = 4.77 \times 10^{-10} \text{ m}$ .
Check the closure at that radius: $v = \sqrt{k e^2/(m_e r)} = 2.18 \times 10^6 \text{ m/s}$ , so $\lambda = h/(m_e v) = 3.33 \times 10^{-10} \text{ m}$ , and $2\pi r = 3.33 \times 10^{-10} \text{ m}$ . One wavelength per lap, exactly as required.
Two caveats worth stating out loud. This derivation goes further than Topic 15.2 asks: nothing in the required course content requires you to solve for a radius. And the value $5.30 \times 10^{-11} \text{ m}$ is not printed anywhere on the AP Physics 2 equation sheet, so it is a result you produce, not a constant you recall. What the derivation does show is that the discreteness is forced by combining the CED's own three relations, not assumed.

(a) The ratio of circumference to de Broglie wavelength is 1.37, not an integer, so $r = 1.0 \times 10^{-10} \text{ m}$ is not an allowed orbit. (b) The smallest allowed radius is $r = \dfrac{h^2}{4\pi^2 m_e k e^2} = 5.30 \times 10^{-11} \text{ m}$ , with allowed radii scaling as $n^2$ .

Frequently asked questions

What is the Bohr model in AP Physics 2?

The Bohr model treats the electron as moving around the nucleus in a circular orbit, with the electric force between the electron and the nucleus acting as the centripetal force. Essential knowledge 15.2.A.3.i states it that way and lists two relevant equations, Coulomb's law F = k q1 q2 / r-squared and the centripetal form F_net = m v-squared / r. The AP Physics 2 CED is explicit that the model is based on classical physics and was the historical representation of the atom that led to the description of the hydrogen atom in terms of discrete energy states.

Why are electron orbits quantized in the Bohr model?

Because of the standing-wave condition in essential knowledge 15.2.A.3.ii: the electron orbit's circumference must be an integer multiple of the electron's de Broglie wavelength. The force balance alone allows any radius, since a smaller orbit simply pairs with a faster electron. Adding the requirement that the electron's wave close on itself after one lap picks out a discrete family of radii indexed by a whole number n, and each of those has its own energy. That is the mechanism behind 15.1.A.5, which says energy and momentum take discrete values for bound systems.

Is the Bohr model correct?

The AP Physics 2 CED treats it as a historical model that works for the hydrogen atom rather than as the current description. Essential knowledge 15.2.A.3 says the Bohr model is based on classical physics and was the historical representation of the atom that led to the description of the hydrogen atom in terms of discrete energy states. The Topic 15.2 boundary statement then limits the analysis and description of electron structure to energy levels and excludes orbitals, orbital shapes, and probability functions, which are exactly the features of the model that replaced it. For exam purposes the Bohr model is the model you use, and its scope is energy levels.

What is the difference between an isotope and an ion?

An isotope differs in the nucleus; an ion differs in the electrons. Essential knowledge 15.2.A.2.ii says the total number of neutrons and protons identifies the isotope of an element, so two isotopes of one element have the same proton count and different neutron counts. Essential knowledge 15.2.A.1.iv says an ion is an atom with a nonzero net electric charge, which happens when the electron count no longer matches the proton count. Changing the proton count instead produces a different element altogether, by 15.2.A.2.

Do you need the formula for hydrogen energy levels on the AP Physics 2 exam?

It is not something you have to supply from memory. The expression E_n equals negative 13.6 eV divided by n squared does not appear in the required course content of Unit 15 and is not printed on the AP Physics 2 equation sheet. It appears once in the whole CED, inside an optional sample instructional activity for Topic 15.3, where teachers are told to tell students that hydrogen's energy levels can be modeled that way. If a question needs numerical energy levels it will provide them, as a diagram or in the stem. The Rydberg formula and the Rydberg constant do not appear in the CED at all.

What does the AP Physics 2 Topic 15.2 boundary statement rule out?

It states that the analysis and description of electron structure is limited to energy levels and will not include such advanced descriptions as orbitals, orbital shapes, or probability functions. In practice that means an atom's electronic structure appears on this exam as a ladder of horizontal energy lines and nothing finer. You will not be asked for s, p or d orbitals, electron configurations, orbital shapes, or probability densities. The related Topic 15.3 boundary statement narrows it further: only energy level diagrams of single-electron atoms will be considered.

Which equations does AP Physics 2 Topic 15.2 use?

Two, and neither is in the Modern Physics group of the equation sheet. Essential knowledge 15.2.A.3.i cites Coulomb's law, F_e = k q1 q2 / r-squared, from the Electricity group, and the centripetal form of Newton's second law, F_net = m v-squared / r, from the Mechanics and Fluids group. Setting them equal for an electron orbiting a proton gives k e-squared / r-squared = m v-squared / r. The de Broglie relation lambda = h/p, cited under Topic 15.1, is what the standing-wave condition in 15.2.A.3.ii then applies to that orbit.