AP Physics 2 · Topic 15.3

Topic 15.3: Emission and Absorption Spectra

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

An atom can only absorb or emit a photon whose energy matches the gap between two of its energy levels. Since every element has its own set of levels, every element has its own set of spectral lines, and that fingerprint is what lets you identify what a distant source is made of.

AP Physics: Unit 15 (topics 15.3 Emission and Absorption Spectra). AP Physics 2 Unit 15, Topic 15.3. One learning objective, 15.3.A, describe the emission or absorption of photons by atoms. Essential knowledge: 15.3.A.1 (energy transfer occurs when photons are absorbed or emitted by an atom, modeled as a system consisting of a nucleus and an electron); 15.3.A.2 (energy can only be absorbed or emitted if the amount corresponds to the energy difference between two atomic energy states) with 15.3.A.2.i (absorb and transition to a higher state), 15.3.A.2.ii (spontaneously emit and move to a lower state), 15.3.A.2.iii (a change in energy state corresponds to a change in the interaction energy between electron and nucleus); 15.3.A.3 (transitions correspond to a photon of a single frequency and therefore a single wavelength); 15.3.A.4 (atoms of each element have a unique set of allowed energy levels and thereby a unique set of absorption and emission frequencies) with 15.3.A.4.i (emission spectrum identifies the elements in a light source), 15.3.A.4.ii (absorption spectrum identifies the elements composing a substance), 15.3.A.4.iii (energy level diagrams represent the energy states of an atom); and 15.3.A.5 (binding energy is the energy required to remove an electron from an atom, causing the atom to become ionized; an atom in the ground state requires the greatest amount of energy to remove the electron). BOUNDARY STATEMENT, printed under this topic: in AP Physics 2, only energy level diagrams of single-electron atoms will be considered. Suggested skills: 1.A, 2.B, 2.C, 3.C. The topic cites no relevant equation of its own; the conversions come from Topic 15.1's E = hf and lambda = c/f plus the printed constant hc = 1240 eV nm. No Rydberg formula or Rydberg constant appears in the CED or on the sheet. E_n = (-13.6 eV)/n^2 appears once in the whole CED, in optional sample instructional activity 1 for this topic, where students view a hydrogen discharge tube through a diffraction grating and identify the red (3 to 2), cyan (4 to 2) and purple (5 to 2) transitions. The Topic 14.4 boundary statement is relevant when discussing colour: students are expected to know the ordering of the electromagnetic spectrum but will not be expected to define exact wavelength ranges.

What Topic 15.3 requires

Topic 15.3 carries one learning objective, 15.3.A: describe the emission or absorption of photons by atoms. Five numbered essential knowledge statements sit under it, and the topic does have a boundary statement.

15.3.A.1 Energy transfer occurs when photons are absorbed or emitted by an atom, which is modeled as a system consisting of a nucleus and an electron.
15.3.A.2 Energy can only be absorbed or emitted by an atom if the amount of energy being absorbed or emitted corresponds to the energy difference between two atomic energy states.
15.3.A.2.i An atom in a given energy state may absorb a photon of the appropriate energy and transition to a higher energy state.
15.3.A.2.ii An atom in an excited energy state may emit a photon of the appropriate energy to spontaneously move to a lower energy state.
15.3.A.2.iii Because an atom is modeled as a system consisting of an electron and a nucleus, a change in the energy state of an atom corresponds to a change in the interaction energy between the electron and the nucleus.
15.3.A.3 Transitions between two energy states of an atom correspond to the absorption or emission of a photon of a single frequency and, therefore, a single wavelength.
15.3.A.4 Atoms of each element have a unique set of allowed energy levels and thereby a unique set of absorption and emission frequencies. The unique set of frequencies determines the element's spectrum.
15.3.A.4.i An emission spectrum can be used to determine the elements in a source of light.
15.3.A.4.ii An absorption spectrum can be used to determine the elements composing a substance by observing what light the substance has absorbed.
15.3.A.4.iii Energy level diagrams are commonly used to visually represent the energy states of an atom.
15.3.A.5 Binding energy is the energy required to remove an electron from an atom, causing the atom to become ionized. An atom in the lowest energy level (ground state) will require the greatest amount of energy to remove the electron from the atom.

Boundary statement: in AP Physics 2, only energy level diagrams of single-electron atoms will be considered.

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.C (justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws).

This topic is the observational half of a pair. Topic 15.2 argues that a bound electron can only occupy certain states. Topic 15.3 is what you see through a spectroscope when that is true. Keep them separate in your head: 15.2 is the model and the mechanism, 15.3 is the evidence and the measurement.

The one rule: only differences

Everything in this topic follows from 15.3.A.2, which is worth carrying in the CED's exact words: energy can only be absorbed or emitted by an atom if the amount of energy being absorbed or emitted corresponds to the energy difference between two atomic energy states.

The word to underline is difference. Not the energy of a level, the gap between two of them. An atom sitting in a state of $-4.5$ eV does not emit a $4.5$ eV photon by virtue of being there. It emits a photon only when it moves, and the photon carries exactly what the move costs.

So the working equation, which the CED never prints as a single line, is

E_{photon} = \left| E_{final} - E_{initial} \right| = hf

with $E = hf$ and $\lambda = c/f$ from the equation sheet doing the conversion, or their combination $E = hc/\lambda$ with the printed $hc = 1240$ eV nm doing it in one step.

The two directions are 15.3.A.2.i and 15.3.A.2.ii, and they are deliberately worded differently:

Absorption (15.3.A.2.i). An atom in a given energy state may absorb a photon of the appropriate energy and transition to a higher energy state. Absorption needs a photon to arrive with the right energy. Arrive with the wrong energy and nothing happens.
Emission (15.3.A.2.ii). An atom in an excited energy state may emit a photon of the appropriate energy to spontaneously move to a lower energy state. Emission needs nothing to arrive. An excited atom drops on its own.

That asymmetry is why an absorption spectrum and an emission spectrum of the same element look like negatives of one another, and it is worth stating explicitly in a 3.C answer.

15.3.A.2.iii supplies the physical picture behind the bookkeeping: because an atom is modeled as a system consisting of an electron and a nucleus, a change in the energy state of an atom corresponds to a change in the interaction energy between the electron and the nucleus. The energy is not stored "in the electron". It is interaction energy of the electron and nucleus considered as a system, exactly as gravitational potential energy belongs to the object-and-Earth system rather than to the object.

One transition, one wavelength

15.3.A.3 is short and its consequences are large: transitions between two energy states of an atom correspond to the absorption or emission of a photon of a single frequency and, therefore, a single wavelength.

A transition does not produce a band, a range, or a spread. It produces a line. That is why a spectrum from an atomic source looks like a set of sharp lines and not like a smooth rainbow, and it is exactly the contrast with Topic 15.4, where a blackbody emits a continuous spectrum. If a question shows you a spectrum, the first thing to read off it is whether it is lines or a continuum, because that tells you whether you are looking at atomic transitions or thermal emission.

The conversion between a level gap and a wavelength is the arithmetic backbone of the topic, and it is worth setting up once, carefully.

You have	You want	Use
Gap in eV	Wavelength in nm	$\lambda = \dfrac{1240 \text{ eV nm}}{\Delta E}$
Wavelength in nm	Gap in eV	$\Delta E = \dfrac{1240 \text{ eV nm}}{\lambda}$
Gap in eV	Gap in J	Multiply by $1.60 \times 10^{-19}$
Gap in J	Frequency in Hz	$f = \dfrac{E}{h}$ with $h = 6.63 \times 10^{-34}$ J s

The two shortcuts are inverses of each other, which gives you a free check: multiply your wavelength in nanometres by your energy in electron volts and you should get 1240 back.

A sense of scale that saves time. A gap of about 2 eV lands near the middle of the visible range; a gap of 12 eV is deep in the ultraviolet; a gap of 0.1 eV is infrared. Note the direction: bigger gap, shorter wavelength. Students reverse this constantly, usually by carrying over the intuition that a bigger number means a bigger wavelength.

One caution about colour language. The CED does expect you to know the ordering of the electromagnetic spectrum, including the colour order red, orange, yellow, green, blue, violet in decreasing wavelength (15.4.A.3.ii in Topic 14.4). But the Topic 14.4 boundary statement says students will not be expected to define exact wavelength ranges within the electromagnetic spectrum. So reason with ordering, not with memorised cut-offs: a 656 nm line is longer than a 486 nm line and therefore redder, and that comparison is defensible.

Emission and absorption: two views of the same fingerprint

15.3.A.4 is the sentence that makes spectroscopy possible: atoms of each element have a unique set of allowed energy levels and thereby a unique set of absorption and emission frequencies, and the unique set of frequencies determines the element's spectrum.

The logic is a chain, and each link is a separate CED statement:

Each element has its own number of protons (15.2.A.2), so its own electric environment for the electron.
So each element has its own set of allowed energy levels (15.3.A.4).
So each element has its own set of level differences.
So each element emits and absorbs its own set of wavelengths (15.3.A.2, 15.3.A.3).
So a measured set of wavelengths identifies the element (15.3.A.4.i and 15.3.A.4.ii).

The two measurement modes:

Emission (15.3.A.4.i). Excite the atoms, let them drop back, and collect what comes out. You see bright lines on a dark background. The CED's use case is stated plainly: an emission spectrum can be used to determine the elements in a source of light. This is the discharge-tube experiment, and it is what the unit's own sample instructional activity has students do with a hydrogen tube and a diffraction grating.

Absorption (15.3.A.4.ii). Shine a continuous spectrum through a cool sample and look at what is missing. You see dark lines on a bright background. An absorption spectrum can be used to determine the elements composing a substance by observing what light the substance has absorbed.

The dark lines sit at the same wavelengths as the bright lines, which is the payoff of the whole topic: both spectra are readouts of the same set of level differences, approached from opposite directions. That symmetry is a ready-made 3.C justification, and the CED's own answer to its essential question "how do we measure things we cannot see".

A subtlety worth knowing but not overclaiming: an absorbing atom does not destroy the energy, it re-emits it, and the reason the line still looks dark is geometric, since the re-emitted photon leaves in some other direction. The CED does not state that, so treat it as background rather than as something to assert on an exam.

Reading an energy level diagram (15.3.A.4.iii and skill 1.A)

15.3.A.4.iii says energy level diagrams are commonly used to visually represent the energy states of an atom, and skill 1.A is listed for this topic, so drawing and reading one is examinable.

The conventions such a diagram uses:

Horizontal lines are allowed energies, usually labelled in electron volts. Vertical position is energy; horizontal position means nothing.
Energies are negative for a bound electron, on the convention that zero energy is a free electron infinitely far from the nucleus and at rest. The most negative line, at the bottom, is the ground state.
The zero line at the top is the ionization limit. Above it the electron is no longer bound.
Arrows are transitions. Downward arrows are emission, upward arrows are absorption, and the length of the arrow is the photon energy.
The gaps get smaller as you go up. The levels crowd together as they approach zero.

The boundary statement fixes what kind of atom you will be shown: in AP Physics 2, only energy level diagrams of single-electron atoms will be considered. Single-electron includes hydrogen and also ions like a doubly ionized lithium atom, which has three protons and one electron. It excludes anything with two or more electrons, which is why no multi-electron spectrum will be put in front of you.

Three counting questions come out of a level diagram, and they are the standard 2.B and 2.C tasks:

How many emission lines can a set of levels produce? From $N$ levels, every pair can in principle produce a transition, so there are $\frac{N(N-1)}{2}$ transitions: 3 for three levels, 6 for four, 10 for five. Watch for the trap in the second worked example below, where two different transitions happen to have the same energy and therefore produce the same line, so the number of distinct wavelengths is smaller than the number of transitions.

Which lines appear in absorption from the ground state? Only transitions that start at the ground state, so $N-1$ of them, all at the short-wavelength end. That is why an absorption spectrum taken through a cool gas usually shows fewer lines than the emission spectrum of the same gas.

Which transition gives the longest wavelength? The smallest gap, which is normally between the two highest levels shown. Longest wavelength always means smallest energy difference.

Binding energy and ionization (15.3.A.5)

The last essential knowledge statement is two sentences and both are examinable: binding energy is the energy required to remove an electron from an atom, causing the atom to become ionized. An atom in the lowest energy level (ground state) will require the greatest amount of energy to remove the electron from the atom.

On a level diagram, the binding energy of a particular state is simply the distance from that line up to the zero line. If a state sits at $-4.5$ eV, then $4.5$ eV frees the electron from it.

The second sentence is the comparison the exam is likely to want, and it follows immediately: the ground state is the most negative line, so it is furthest from zero, so it has the largest gap to climb. Excite the atom first and you have already paid part of the bill, and the remaining binding energy is smaller. That is a clean skill 2.C comparison, and it also connects to 15.2.A.1.iv, since the result of removing the electron is an ion, an atom with a nonzero net electric charge.

Two distinctions worth keeping straight, because the vocabulary in this area is crowded:

Binding energy of an electron in an atom (15.3.A.5) is what this topic means. It is about pulling an electron away from a nucleus.
Binding energy of a nucleus turns up in Topic 15.7, where 15.7.A.8 says nuclear fission may occur spontaneously or may require an energy input depending on the binding energy of the nucleus. That is about pulling nucleons apart, an entirely different scale of energy.
Work function (15.5.A.3.i, in Topic 15.5) is the minimum energy required to emit an electron from atoms in a material. It is the same idea applied to a bulk solid rather than to an isolated atom, and the Topic 15.5 boundary statement says its values will be provided on the exam.

Same concept, three settings, three names. If an answer needs one of them, name which.

What is not printed, and what a question will hand you

Topic 15.3 cites no equation of its own in the CED's Required Course Content. The conversions it needs come from Topic 15.1: $E = hf$ and $\lambda = c/f$ , both printed in the Modern Physics group of the AP Physics 2 equation sheet, plus the printed constant $hc = 1240$ eV nm from the Table of Information.

What is not available anywhere:

No Rydberg formula and no Rydberg constant. Neither appears in the AP Physics 2 CED or on its equation sheet. If you learned the $1/\lambda = R(1/n_1^2 - 1/n_2^2)$ form elsewhere, it is not the route this course expects, and you will not be handed $R$ .
No general 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 entire CED, in the Unit 15 Sample Instructional Activities table, attached to Topic 15.3. The activity has teachers show a hydrogen discharge tube, hand out a diffraction grating so students can see the red, cyan and purple lines, tell students that the energy levels of hydrogen can be modeled that way, and have them work out which transitions they are seeing (red 3 to 2, cyan 4 to 2, purple 5 to 2). The page carrying that table states the activities are optional and that teachers do not need to use them.
No table of energy levels for any element.

The practical rule: the numbers come with the question. A Topic 15.3 problem gives you a level diagram, or a list of level energies, or a formula in the stem. Your job is the differences, the conversions, and the reasoning about which transitions can occur. That is the same arrangement the Topic 15.5 boundary statement makes explicit for work functions, and it is a deliberate design choice across the whole unit.

The first worked example below runs the CED's own optional activity end to end, because it is the CED's own illustration of the topic and the three wavelengths it produces are checkable against the colours the CED itself names.

How Topic 15.3 is tested

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

The four listed skills map onto four recognisable question types.

1.A, represent. Draw or complete an energy level diagram, or mark on it the transition responsible for a given line. Get the arrow direction right: down is emission, up is absorption.

2.B, calculate. Turn a level gap into a wavelength or a frequency, or the reverse. The whole difficulty here is unit discipline, and the reliable habit is to work in electron volts and nanometres with $hc = 1240$ eV nm, converting to joules only if the question asks for joules.

2.C, compare. Two transitions, two elements, two states. Which photon has the greater energy, the longer wavelength, the higher frequency. Which state has the larger binding energy. These are answerable without a calculator once you have internalised that a larger gap means a shorter wavelength.

3.C, justify with evidence. Explain why a given spectrum identifies a given element, or why a photon of a stated energy passes straight through an atom without being absorbed. The evidence you are pointing at is 15.3.A.2: the energy has to correspond to a difference between two states, and if no pair of states differs by that amount, nothing happens.

A checklist before you leave the topic. Can you state, in the CED's terms, the one condition under which an atom absorbs or emits? Can you convert a gap in electron volts to a wavelength in nanometres in one step, and check the answer by multiplying back to 1240? Can you count the transitions available from a four-level diagram, and notice when two of them coincide? Can you say why the ground state has the largest binding energy? Can you say what the boundary statement excludes?

Where this sits in the unit: 15.2 gives you levels, 15.3 turns them into light, and 15.4 supplies the contrasting case of a continuous spectrum with no lines in it at all.

The CED's hydrogen discharge tube, worked through

Sample instructional activity 1 for Unit 15 has students view a hydrogen discharge tube through a diffraction grating, see red, cyan and purple lines, and use the model $E_n = \frac{-13.6 \text{ eV}}{n^2}$ to identify the transitions. Find the wavelengths of the $n = 3 \to 2$ , $n = 4 \to 2$ and $n = 5 \to 2$ transitions, and check that their order matches the colours.

First, note the status of that formula. It is not on the equation sheet and it is not in the required course content. It comes from the stem of this activity, exactly the way a real exam question would supply it. Use it because you were given it.
Evaluate the levels you need: $E_2 = \dfrac{-13.6}{4} = -3.40 \text{ eV}$ , $E_3 = \dfrac{-13.6}{9} = -1.511 \text{ eV}$ , $E_4 = \dfrac{-13.6}{16} = -0.850 \text{ eV}$ , $E_5 = \dfrac{-13.6}{25} = -0.544 \text{ eV}$ .
Apply 15.3.A.2, the photon energy is the difference between two states. For $3 \to 2$ : $\Delta E = E_3 - E_2 = -1.511 - (-3.40) = 1.889 \text{ eV}$ . Then $\lambda = \dfrac{1240 \text{ eV nm}}{1.889 \text{ eV}} = 656 \text{ nm}$ .
For $4 \to 2$ : $\Delta E = -0.850 - (-3.40) = 2.55 \text{ eV}$ , so $\lambda = \dfrac{1240}{2.55} = 486 \text{ nm}$ .
For $5 \to 2$ : $\Delta E = -0.544 - (-3.40) = 2.856 \text{ eV}$ , so $\lambda = \dfrac{1240}{2.856} = 434 \text{ nm}$ .
Check the ordering against the colours. The CED's activity says red is 3 to 2, cyan is 4 to 2, and purple is 5 to 2. Our wavelengths run $656 > 486 > 434$ nm, decreasing in that same order, and 14.4.A.3.ii gives the colour order in decreasing wavelength as red, orange, yellow, green, blue, violet. Longest goes with reddest. The assignment is consistent.
Check one line the other way as well, since the check is free: $(656 \text{ nm})(1.889 \text{ eV}) = 1239$ , which is $1240$ to the precision of the rounding. Wavelength in nanometres times energy in electron volts should always return $1240$ .
One observation about the pattern. All three lines end on $n = 2$ , not on the ground state $n = 1$ . Transitions down to $n = 1$ from these levels are much larger jumps, $12.1$ eV or more, which puts them at $103$ nm and shorter, well past violet in the ordering of 14.4.A.3.i. The visible lines of hydrogen are the ones that stop one rung up.

$3 \to 2$ gives $656 \text{ nm}$ , $4 \to 2$ gives $486 \text{ nm}$ , and $5 \to 2$ gives $434 \text{ nm}$ . The wavelengths decrease in the order red, cyan, purple, matching the CED's assignment and the colour ordering in 14.4.A.3.ii.

Counting the lines from a four-level diagram

A single-electron atom has four energy levels: $E_1 = -8.0 \text{ eV}$ (ground state), $E_2 = -4.5 \text{ eV}$ , $E_3 = -2.0 \text{ eV}$ , and $E_4 = -1.0 \text{ eV}$ . (a) How many downward transitions are possible, and what are their photon energies? (b) How many distinct wavelengths appear in the emission spectrum? (c) Which transition produces the longest wavelength?

(a) With four levels, every pair gives one downward transition, so there are $\dfrac{4 \times 3}{2} = 6$ of them. Take each difference, using 15.3.A.2.
$4 \to 3$ : $-1.0 - (-2.0) = 1.0 \text{ eV}$ . $4 \to 2$ : $-1.0 - (-4.5) = 3.5 \text{ eV}$ . $4 \to 1$ : $-1.0 - (-8.0) = 7.0 \text{ eV}$ .
$3 \to 2$ : $-2.0 - (-4.5) = 2.5 \text{ eV}$ . $3 \to 1$ : $-2.0 - (-8.0) = 6.0 \text{ eV}$ . $2 \to 1$ : $-4.5 - (-8.0) = 3.5 \text{ eV}$ .
(b) Look at the list of energies: $1.0$ , $3.5$ , $7.0$ , $2.5$ , $6.0$ , $3.5$ eV. Two of them are identical. The $4 \to 2$ jump and the $2 \to 1$ jump both happen to span $3.5 \text{ eV}$ , so by 15.3.A.3 they produce photons of the same single frequency and therefore the same single wavelength. Six transitions, but only five distinct wavelengths.
Convert them: $\lambda = 1240/\Delta E$ gives $1240 \text{ nm}$ for $1.0$ eV, $496 \text{ nm}$ for $2.5$ eV, $354 \text{ nm}$ for $3.5$ eV, $207 \text{ nm}$ for $6.0$ eV, and $177 \text{ nm}$ for $7.0$ eV.
(c) The longest wavelength comes from the smallest energy difference, which is the $1.0 \text{ eV}$ gap between the two top levels, giving $1240 \text{ nm}$ . Note how far that is from the $177 \text{ nm}$ line at the other end: the same atom emits across a factor of seven in wavelength.
A caution on the coincidence in part (b). It is a property of these particular level values, not a general rule, and it is a favourite way to make a counting question harder than it looks. Always compare the energies before counting lines, rather than assuming that $N(N-1)/2$ transitions give $N(N-1)/2$ wavelengths.

(a) Six transitions, at $1.0$ , $2.5$ , $3.5$ , $3.5$ , $6.0$ and $7.0 \text{ eV}$ . (b) Five distinct wavelengths, $1240$ , $496$ , $354$ , $207$ and $177 \text{ nm}$ , because $4 \to 2$ and $2 \to 1$ coincide at $3.5 \text{ eV}$ . (c) The $4 \to 3$ transition, at $1240 \text{ nm}$ .

Absorption from the ground state, and binding energy

Use the same atom, with levels at $-8.0$ , $-4.5$ , $-2.0$ and $-1.0 \text{ eV}$ . The atom is in its ground state. (a) Which of the following incident photons are absorbed: $2.5 \text{ eV}$ , $3.5 \text{ eV}$ , $5.0 \text{ eV}$ , $6.0 \text{ eV}$ ? (b) State the binding energy of the atom in the ground state and in the first excited state, and say which is greater. (c) What wavelength of light would just ionize the atom from its ground state?

(a) Apply 15.3.A.2.i: an atom in a given energy state may absorb a photon of the appropriate energy and transition to a higher energy state. Starting from $-8.0 \text{ eV}$ , the only available destinations are $-4.5$ , $-2.0$ and $-1.0 \text{ eV}$ , so the only absorbable energies are $3.5$ , $6.0$ and $7.0 \text{ eV}$ .
So the $3.5 \text{ eV}$ and $6.0 \text{ eV}$ photons are absorbed. The $2.5 \text{ eV}$ photon is not: $2.5 \text{ eV}$ is a real gap in this atom, between levels 2 and 3, but the atom is not in level 2, so that gap is not available from where it is. The $5.0 \text{ eV}$ photon is not absorbed either, because no pair of levels differs by $5.0 \text{ eV}$ at all.
That $2.5 \text{ eV}$ case is the one worth remembering. Whether a photon is absorbed depends on the state the atom is in, not only on the set of gaps the atom has. This is also why an absorption spectrum through a cool gas shows fewer lines than the emission spectrum of the same gas: cool atoms are nearly all in the ground state.
(b) By 15.3.A.5, binding energy is the energy required to remove the electron, which on a level diagram is the distance from the state up to zero. From the ground state at $-8.0 \text{ eV}$ the binding energy is $8.0 \text{ eV}$ . From the first excited state at $-4.5 \text{ eV}$ it is $4.5 \text{ eV}$ .
The ground-state value is the greater of the two, which is exactly what 15.3.A.5 asserts: an atom in the lowest energy level will require the greatest amount of energy to remove the electron. Exciting the atom first has already supplied $3.5 \text{ eV}$ of the $8.0 \text{ eV}$ needed.
(c) Just ionizing from the ground state needs a photon of $8.0 \text{ eV}$ , so $\lambda = \dfrac{1240 \text{ eV nm}}{8.0 \text{ eV}} = 155 \text{ nm}$ . Check by multiplying back: $(155)(8.0) = 1240$ .
Compare that with ionizing from the first excited state, which needs only $4.5 \text{ eV}$ , or $\lambda = 1240/4.5 = 276 \text{ nm}$ . Longer wavelength, lower energy, because the electron was already most of the way out. Once the electron is removed the atom carries a nonzero net charge and is, by 15.2.A.1.iv, an ion.

(a) The $3.5 \text{ eV}$ and $6.0 \text{ eV}$ photons are absorbed; the $2.5 \text{ eV}$ and $5.0 \text{ eV}$ photons are not. (b) Binding energy is $8.0 \text{ eV}$ from the ground state and $4.5 \text{ eV}$ from the first excited state, so the ground state requires more, as 15.3.A.5 states. (c) $155 \text{ nm}$ .

Frequently asked questions

Why do atoms only emit certain wavelengths of light?

Because an atom's energy is restricted to a set of allowed levels, and a photon can only carry away the difference between two of them. Essential knowledge 15.3.A.2 in the AP Physics 2 CED states that energy can only be absorbed or emitted by an atom if the amount of energy corresponds to the energy difference between two atomic energy states, and 15.3.A.3 adds that each such transition produces a photon of a single frequency and therefore a single wavelength. Since the set of level differences is a finite list, the set of emitted wavelengths is a finite list too, which is what a line spectrum looks like.

What is the difference between an emission spectrum and an absorption spectrum?

An emission spectrum is bright lines on a dark background, produced when excited atoms drop to lower energy states and release photons. An absorption spectrum is dark lines on a bright background, produced when a continuous spectrum passes through a substance and the atoms take out the photons whose energies match their level differences. The lines fall at the same wavelengths in both cases because both are readouts of the same set of energy gaps. The AP Physics 2 CED gives each a use: 15.3.A.4.i says an emission spectrum can be used to determine the elements in a source of light, and 15.3.A.4.ii says an absorption spectrum can be used to determine the elements composing a substance.

How do you find the wavelength of a photon from an energy level diagram?

Take the difference between the two levels in electron volts, then divide 1240 by it to get the wavelength in nanometres. The AP Physics 2 Table of Information prints hc = 1240 eV nm, and combining the printed equations E = hf and lambda = c/f gives E = hc / lambda. So a transition across a 2.5 eV gap gives 1240 / 2.5 = 496 nm. Check the arithmetic by multiplying back: wavelength in nanometres times energy in electron volts should always come to 1240. A bigger energy gap always means a shorter wavelength.

What is binding energy in AP Physics 2 Topic 15.3?

Essential knowledge 15.3.A.5 defines binding energy as the energy required to remove an electron from an atom, causing the atom to become ionized, and adds that an atom in the lowest energy level, the ground state, will require the greatest amount of energy to remove the electron. On an energy level diagram the binding energy of a state is the distance from that level up to the zero line, so a state at negative 4.5 eV has a binding energy of 4.5 eV. Do not confuse it with nuclear binding energy, which appears in Topic 15.7, or with the work function of a material, which appears in Topic 15.5.

Is the Rydberg formula used in AP Physics 2?

No. Neither the Rydberg formula nor the Rydberg constant appears anywhere in the AP Physics 2 course and exam description, and neither is printed on the AP Physics 2 equation sheet. Topic 15.3 works from energy differences instead: subtract two level energies to get the photon energy, then convert with the printed hc = 1240 eV nm. If a question needs hydrogen's levels specifically, it supplies them. The expression E_n equals negative 13.6 eV over n squared appears once in the whole CED, in an optional sample instructional activity attached to this topic.

What does the Topic 15.3 boundary statement say?

It 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 qualifies. What it rules out is any atom with two or more electrons, so no multi-electron spectra and no diagrams whose structure depends on electron-electron interactions. It pairs with the Topic 15.2 boundary statement, which limits the description of electron structure to energy levels and excludes orbitals, orbital shapes and probability functions.

How many spectral lines can an atom with four energy levels produce?

Up to six, since each pair of levels gives one transition and four levels have six pairs, following N(N-1)/2. But the number of distinct wavelengths can be smaller, because two different transitions can span the same energy gap and therefore produce the same wavelength. For levels at negative 8.0, negative 4.5, negative 2.0 and negative 1.0 eV, the transitions from level 4 to level 2 and from level 2 to level 1 both span 3.5 eV, so the six transitions produce only five distinct lines. Compare the energy differences before counting lines.