15.3 Emission and Absorption Spectra

1. AThe atom's energy increases by exactly the energy of the absorbed photon. Correct
   Absorption transfers the photon's energy to the atom, the system of nucleus and electron, in a single event. The photon no longer exists, and the atom's energy rises by the photon's energy, hf.
2. BThe atom's energy increases by part of the photon's energy; a weaker photon carries on.
   A student who thinks an atom can take just the energy it needs picks this. An atom absorbs a whole photon or none of it: the photon is absorbed only if its energy matches a transition (or can ionize the atom).
3. CThe atom's energy increases by an amount that depends on how bright the beam is.
   A student who links brightness with photon energy picks this. The energy of each photon is hf, set by the frequency; a brighter beam has more photons, not more energetic ones.
4. DThe atom's energy increases gradually for as long as the beam shines on it.
   A student who pictures light pouring energy into the atom continuously picks this. The energy arrives in one step, when the photon is absorbed; it does not build up over time.

CED 15.3.A.1 · Read this in Fix

1. AIt now belongs to the electron alone, since the nucleus has no part in the energy.
   A student who attaches the energy levels to the electron alone picks this. The energy states are states of the electron–nucleus system; the interaction energy belongs to the pair.
2. BIt is now kinetic energy of the electron, which moves faster in the n = 2 state.
   A student who expects more energy to mean more speed picks this. In the Bohr model the electron in the n = 2 orbit is farther from the nucleus and moves more slowly; the energy has gone into the electron–nucleus interaction.
3. CIt is now energy of the electron–nucleus system, which is set by how the two interact. Correct
   The atom is modeled as a system of an electron and a nucleus. Moving to a higher energy state means an increase in the interaction energy between the electron and the nucleus, and the photon's energy supplies that increase.
4. DIt is still in the photon, which is stored in the atom until it is given back.
   A student who thinks an absorbed photon survives inside the atom picks this. The photon ceases to exist when it is absorbed; only its energy remains, as energy of the atom.

CED 15.3.A.2.iii · Read this in Fix

1. A2.00 Correct
   The photon energies are the level differences, 6.00 eV for X and 3.00 eV for Y. The wavelength is inversely proportional to the photon energy (λ = hc/ΔE), so half the energy means twice the wavelength: λ_Y/λ_X = 2.00.
2. B0.50
   A student who thinks a larger energy difference gives a longer wavelength picks this, taking λ proportional to ΔE. Since λ = hc/ΔE, the transition with half the energy difference has twice the wavelength.
3. C2.50
   A student who takes each photon energy to be the magnitude of the energy of the level the atom ends in, 10.0 eV for X and 4.00 eV for Y, picks this. The photon energies are the differences between the levels, 6.00 eV and 3.00 eV.
4. D0.67
   A student who measures every transition from the ground state picks this, using 9.00 eV for Y. Transition Y goes from −1.00 eV to −4.00 eV, so its energy is 3.00 eV.

Working ΔE_X = −4.00 eV − (−10.0 eV) = 6.00 eV; ΔE_Y = −1.00 eV − (−4.00 eV) = 3.00 eV. λ = hc/ΔE, so λ is inversely proportional to ΔE: λ_Y/λ_X = ΔE_X/ΔE_Y = 6.00/3.00 = 2.00. Errors: λ taken as proportional to ΔE → 0.50; photon energy = magnitude of the final level's energy (10.0 eV and 4.00 eV) → 10.0/4.00 = 2.50; energies measured from the ground state (6.00 eV and 9.00 eV) → 0.67.

CED 15.3.A.3 · Read this in Fix

1. AThe distance between the electron and the nucleus in that state
   A student who reads the diagram as a picture of orbits picks this. The vertical axis is energy, not distance; the spacing between lines shows energy differences.
2. BThe energy of the atom, the system of electron and nucleus, in that state Correct
   An energy-level diagram plots energy vertically. Each line is one allowed state of the atom, the electron–nucleus system, drawn at the height of that state's energy; the vertical spacing between lines shows energy differences.
3. CThe energy that belongs to the electron alone, apart from the nucleus
   A student who attaches the energy levels to the electron alone picks this. The states are states of the electron–nucleus system; the energy is interaction energy of the pair.
4. DThe number of electrons that the state can hold, as in a shell
   A student who carries over the shell model from chemistry picks this. A single-electron atom has one electron, and each line is a possible energy state of the whole atom.

CED 15.3.A.4.iii · Read this in Fix

1. A3.40 eV Correct
   The binding energy is the energy needed to remove the electron, taking the atom from its present state to ionization at 0 eV: 0 − (−3.40 eV) = 3.40 eV.
2. B13.6 eV
   A student who thinks an atom has one fixed binding energy picks this. 13.6 eV is the binding energy of the ground state; an atom already in n = 2 needs only 3.40 eV to be ionized.
3. C2.55 eV
   A student who treats the highest level listed, n = 4, as ionization picks this: −0.850 eV − (−3.40 eV) = 2.55 eV. There are more bound levels above n = 4; the electron is removed only when the energy reaches 0 eV.
4. D10.2 eV
   A student who confuses the binding energy with the energy needed to reach n = 2 from the ground state picks this: −3.40 eV − (−13.6 eV) = 10.2 eV. The binding energy is measured from n = 2 up to 0 eV.

Working Binding energy from n = 2 = 0 − E₂ = 0 − (−3.40 eV) = 3.40 eV. (Ground-state value 13.6 eV; to n = 4: 2.55 eV; n = 1 → n = 2: 10.2 eV.)

CED 15.3.A.5 · Read this in Fix

1. AThe atom absorbs 4.0 eV, and a photon of energy 0.5 eV continues on.
   A student who thinks an atom can take the part of a photon's energy it needs picks this. A photon is absorbed whole or not at all, and 4.5 eV does not equal any energy difference from the ground state.
2. BThe photon is not absorbed, and it continues on with energy 4.5 eV. Correct
   From the ground state (−6.0 eV) the atom can absorb only 0.5 eV (to −5.5 eV), 4.0 eV (to −2.0 eV), 5.0 eV (to −1.0 eV), or at least 6.0 eV (ionization). A 4.5 eV photon matches none of these, so it is not absorbed and passes on unchanged.
3. CThe photon is absorbed, and the atom's energy becomes −1.5 eV.
   A student who thinks an atom's energy can change by any amount picks this: −6.0 eV + 4.5 eV = −1.5 eV. There is no state at −1.5 eV, so the atom cannot end up there, and the photon is not absorbed.
4. DThe photon is absorbed, and the atom ends in the −1.0 eV level.
   A student who uses any gap between two of the listed levels picks this: 4.5 eV is the gap from −5.5 eV to −1.0 eV. An atom in the ground state can use only gaps that start at −6.0 eV: 0.5 eV, 4.0 eV, or 5.0 eV, or at least 6.0 eV to ionize it.

Working Ground state −6.0 eV. Allowed absorptions: to −5.5 eV needs 0.5 eV; to −2.0 eV needs 4.0 eV; to −1.0 eV needs 5.0 eV; ionization needs at least 6.0 eV. 4.5 eV matches none → not absorbed; the photon continues with 4.5 eV. (The 4.5 eV gap from −5.5 eV to −1.0 eV is not available from the ground state.)

CED 15.3.A.2 · Read this in Fix

1. A3.0 eV
   A student who uses any gap on the diagram picks this: 3.0 eV is the gap between n = 2 and n = 3. An atom in the ground state can use only gaps that start at the ground state.
2. B7.0 eV
   A student who reads the energy of the level to be reached as the photon energy picks this, taking 7.0 eV for n = 2. Reaching n = 2 from the ground state takes −7.0 eV − (−12.0 eV) = 5.0 eV.
3. C8.0 eV Correct
   From the ground state, −12.0 eV, the atom can absorb a photon whose energy takes it exactly to another level: 5.0 eV (to −7.0 eV), 8.0 eV (to −4.0 eV), or 9.5 eV (to −2.5 eV), or to a higher level not shown (a photon energy between 9.5 eV and 12.0 eV could match one), or at least 12.0 eV to ionize it. Of the options, only 8.0 eV matches.
4. D6.0 eV
   A student who thinks an atom can absorb any amount of energy picks this; the atom would end at −6.0 eV. No state has that energy, and 6.0 eV is less than the 12.0 eV needed for ionization, so the photon is not absorbed.

Working Ground state −12.0 eV. Differences: n = 2: 5.0 eV; n = 3: 8.0 eV; n = 4: 9.5 eV; ionization ≥ 12.0 eV. Only 8.0 eV is among the options.

CED 15.3.A.2.i · Read this in Fix

1. A4.25 eV
   A student who adds the magnitudes of the two energies picks this: 0.850 eV + 3.40 eV = 4.25 eV. Subtracting with the signs gives (−0.850 eV) − (−3.40 eV) = 2.55 eV.
2. B3.40 eV
   A student who reads the energy of the final level as the photon energy picks this. That value is measured from 0 eV, not from the n = 4 level; the photon energy is the difference between the two levels.
3. C12.8 eV
   A student who measures every transition from the ground state picks this: (−0.850 eV) − (−13.6 eV) = 12.75 eV. The atom ends in n = 2, not n = 1, so the difference is taken to the n = 2 level.
4. D2.55 eV Correct
   The photon carries away the energy difference between the two states: E = E₄ − E₂ = (−0.850 eV) − (−3.40 eV) = 2.55 eV.

Working E_photon = E₄ − E₂ = −0.850 eV − (−3.40 eV) = 2.55 eV. (Sum of magnitudes 4.25 eV; |E₂| = 3.40 eV; E₄ − E₁ = 12.75 eV ≈ 12.8 eV.)

CED 15.3.A.2.ii · Read this in Fix

1. AIn transition 1, and in no other
   A student who has exchanged absorption and emission picks this. Transition 1 goes up, so the atom must take in energy, for example by absorbing a photon; the three downward transitions emit photons.
2. BIn transitions 2, 3, and 4 only Correct
   An atom emits a photon when it moves to a lower energy state, carrying away the energy difference. Transitions 2 (n = 3 to n = 2), 3 (n = 4 to n = 1), and 4 (n = 2 to n = 1) all go down; transition 1 goes up, which requires absorbing energy.
3. CIn transitions 3 and 4 only
   A student who thinks an excited atom emits only when it returns to the ground state picks this. Transition 2, from n = 3 to n = 2, is also downward, so it emits a photon too.
4. DIn all four of the transitions
   A student who thinks every change of state gives off light picks this. Only downward transitions emit photons; transition 1, upward, requires the atom to absorb energy.

Working Emission ↔ downward transition. 1: n = 1 → n = 3 (up, absorption). 2: n = 3 → n = 2, 3: n = 4 → n = 1, 4: n = 2 → n = 1 (all down, emission).

CED 15.3.A.2.ii · Read this in Fix

1. Aλ_X > λ_Z > λ_Y Correct
   Each transition gives a photon of one wavelength, λ = hc/ΔE. The energy differences are X: 3.0 eV, Y: 9.0 eV, Z: 6.0 eV. The smallest energy difference gives the longest wavelength, so λ_X > λ_Z > λ_Y.
2. Bλ_Y > λ_Z > λ_X
   A student who thinks a larger energy difference gives a longer wavelength picks this. Wavelength is inversely proportional to photon energy, so the largest gap, Y, gives the shortest wavelength.
3. Cλ_X = λ_Y = λ_Z
   A student who thinks an element emits a single wavelength picks this. Each pair of levels gives its own wavelength, and the three energy differences here are all different.
4. Dλ_X > λ_Y = λ_Z
   A student who takes the photon energy as the energy of the final level picks this: 4.0 eV for X and 10.0 eV for both Y and Z. The photon energy is the difference between the two levels, which is 9.0 eV for Y and 6.0 eV for Z.

Working ΔE_X = −1.0 − (−4.0) = 3.0 eV; ΔE_Y = −1.0 − (−10.0) = 9.0 eV; ΔE_Z = −4.0 − (−10.0) = 6.0 eV. λ = hc/ΔE: λ_X = 413 nm > λ_Z = 207 nm > λ_Y = 138 nm.

CED 15.3.A.3 · Read this in Fix

1. AThe atoms can change energy by any amount, though some amounts occur more often.
   A student who thinks an atom's energy can change by any amount picks this. Changes of any size would give light at every wavelength, a continuous band, not separate lines with nothing between them.
2. BThe gas is a mixture, and each line comes from an element that emits one wavelength.
   A student who thinks each element emits a single wavelength picks this. One element can produce several lines, one for each pair of its energy levels, so several lines do not show a mixture.
3. CThe photons have fixed energies because the glowing tube's brightness is fixed.
   A student who links photon energy with brightness picks this. The energy of each photon is set by the transition that produced it; a brighter tube would give brighter lines at the same wavelengths.
4. DOnly particular energy changes are possible, and each one produces a single wavelength. Correct
   Each transition between two energy states gives photons of a single frequency and so a single wavelength. A few sharp lines therefore show that the atoms change energy only by a few particular amounts.

CED 15.3.A.3 · Read this in Fix

1. AThe spectra are identical, since the energy states belong to the single electron alone.
   A student who thinks the energy states belong to the electron alone picks this. The states depend on the interaction between the electron and the nucleus, and the two nuclei are different.
2. BEach spectrum is a single line, at a different wavelength for hydrogen and for He⁺.
   A student who thinks each kind of atom emits one wavelength picks this. Each has many energy levels, and every pair of levels gives a line, so each spectrum has many lines.
3. CThe spectra differ, since the two different nuclei give different sets of energy levels. Correct
   An atom's energy states are states of the electron–nucleus system, set by the electric interaction between them. The He⁺ nucleus has twice the charge, so He⁺ has a different set of energy levels, and so a different set of emission and absorption frequencies.
4. DThe spectra differ, since the heavier He⁺ nucleus pulls harder on the electron by gravity.
   A student who thinks gravity holds the electron picks this. The spectra do differ, but because of the electric interaction with a nucleus of different charge; gravity is about 10³⁹ times weaker and plays no part.

CED 15.3.A.4 · Read this in Fix

1. AIt contains each of the elements A, B, and C.
   A student who accepts a single matching line picks this. B shares only the 660 nm line with the lamp; its lines at 430 nm and 580 nm are missing, so B is not present.
2. BIt contains elements A and C, but not B. Correct
   Every line of A (450, 520, 610 nm) and every line of C (470, 660 nm) appears in the lamp's spectrum, and together they account for all five of its lines. B's lines at 430 nm and 580 nm are missing, so B is not present; its 660 nm line is explained by C.
3. CIt contains element A, and no other element.
   A student who assumes a lamp contains a single element picks this, choosing the best match. A explains only three of the five lines; the lines at 470 nm and 660 nm show that C is present as well.
4. DIt contains none of the three elements.
   A student who thinks a mixture of gases gives a new pattern picks this. Each element emits its own lines whatever it is mixed with, so the lamp's spectrum is the lines of A and C together.

Working Lamp lines: 450, 470, 520, 610, 660 nm. A (450, 520, 610) all present; C (470, 660) all present; B (430, 580, 660): 430 and 580 absent → B absent. A + C account for all five lines.

CED 15.3.A.4.i · Read this in Fix

1. AAt 6.0 eV and at 8.0 eV only Correct
   Nearly all the atoms are in the ground state, −10.0 eV, so they can absorb only photons that take them from n = 1 to a higher level: 6.0 eV (to n = 2) and 8.0 eV (to n = 3). Those photons are removed from the light, giving dark lines at 6.0 eV and 8.0 eV.
2. BAt 2.0 eV, 6.0 eV, and 8.0 eV
   A student who uses every gap on the diagram picks this. The 2.0 eV gap, from n = 2 to n = 3, can be used only by atoms already in n = 2, and almost none are in a cool gas.
3. CAt 6.0 eV, the step up to n = 2
   A student who thinks an atom can move only to the next level picks this. An atom in the ground state can also absorb 8.0 eV and go directly from n = 1 to n = 3.
4. DAt 2.0 eV and 4.0 eV only
   A student who reads each level's energy as the energy needed to reach it picks this. Reaching n = 2 from the ground state takes −4.0 eV − (−10.0 eV) = 6.0 eV, not 4.0 eV, and reaching n = 3 takes 8.0 eV, not 2.0 eV.

Working Cool gas: absorption only from n = 1 (−10.0 eV). n = 1 → n = 2: 6.0 eV; n = 1 → n = 3: 8.0 eV. n = 2 → n = 3 (2.0 eV) needs atoms in n = 2, which are almost absent. Ionization needs at least 10.0 eV.

CED 15.3.A.4.ii · Read this in Fix

1. AThe cloud has no Q in it, since Q's wavelengths are missing from the emerging light.
   A student who reads dark lines as missing elements picks this. The light at those wavelengths is missing because Q's atoms in the cloud absorbed it; the dark lines show that Q is present.
2. BThe cloud contains some element other than Q, since Q absorbs at other wavelengths.
   A student who thinks an element absorbs and emits at different wavelengths picks this. Absorption and emission use the same energy differences, so Q's dark lines fall at wavelengths among its bright lines.
3. CThe source gives off no light at those wavelengths, so the cloud plays no part.
   A student who blames the source for the dark lines picks this. The source gives a continuous spectrum, light at every wavelength; the dark lines appear because the gas in the cloud absorbs particular wavelengths.
4. DThe cloud contains element Q, whose atoms absorbed photons of those wavelengths. Correct
   An element absorbs at wavelengths that also appear among its emission lines, because both processes involve the same pairs of energy states. Dark lines exactly at Q's wavelengths show that atoms of Q in the cloud absorbed those photons.

CED 15.3.A.4.ii · Read this in Fix

1. AThree wavelengths
   A student who thinks an atom can move only between neighboring levels picks this, counting 4→3, 3→2, and 2→1. Transitions such as 4→1 or 3→1 skip levels and give lines too.
2. BFour wavelengths
   A student who expects one line for each level picks this. Lines come from pairs of levels, and four levels make six pairs.
3. CSix wavelengths Correct
   A photon is emitted in a downward transition between any two levels, so each pair of levels gives one wavelength: 4→3, 4→2, 4→1, 3→2, 3→1, and 2→1. The six energy differences (1.0, 3.0, 6.5, 2.0, 5.5, and 3.5 eV) are all different, so there are six wavelengths.
4. DOne wavelength
   A student who thinks each element emits a single characteristic wavelength picks this. Each pair of levels gives its own line: four levels make six pairs, and here the six energy differences are all different, so the gas emits six wavelengths.

Working Pairs of 4 levels: 4 × 3 / 2 = 6. Energy differences: 4→3: 1.0 eV; 4→2: 3.0 eV; 4→1: 6.5 eV; 3→2: 2.0 eV; 3→1: 5.5 eV; 2→1: 3.5 eV — all different, so six wavelengths.

CED 15.3.A.4.iii · Read this in Fix

1. Aλ + λ′
   A student who thinks a larger energy difference gives a proportionally longer wavelength picks this, adding the wavelengths as if they were energies. A photon's energy is hc/λ, so the 3 → 1 photon, with the largest energy, has the shortest wavelength: the energies add, not the wavelengths.
2. Bλ
   A student who measures the energy of every transition from the ground state picks this, taking the photon of wavelength λ to carry E₃ − E₁. That photon is emitted in the 3 → 2 transition, so its energy is E₃ − E₂, which is less than E₃ − E₁.
3. Cλλ′/(λ + λ′) Correct
   The energy differences add: E₃ − E₁ = (E₃ − E₂) + (E₂ − E₁) = hc/λ + hc/λ′. So 1/λ₃₁ = 1/λ + 1/λ′, which gives λ₃₁ = λλ′/(λ + λ′). This is shorter than both λ and λ′, as the largest energy difference requires.
4. Dλ′
   A student who takes a photon's energy to be the magnitude of the energy of the state the atom ends in picks this: the 3 → 1 and 2 → 1 photons both end in state 1, so they would have the same wavelength. The photon's energy is the difference between the two states, which is larger for 3 → 1 than for 2 → 1.

Working E₃ − E₂ = hc/λ and E₂ − E₁ = hc/λ′. E₃ − E₁ = (E₃ − E₂) + (E₂ − E₁) = hc(1/λ + 1/λ′) = hc(λ + λ′)/(λλ′). λ₃₁ = hc/(E₃ − E₁) = λλ′/(λ + λ′). Check: λ = 400 nm, λ′ = 600 nm → 240 nm, shorter than both. Errors: wavelengths added as if proportional to energy → λ + λ′; the 3 → 2 photon's energy measured from the ground state (E₃ − E₁) → λ; photon energy = magnitude of the final state's energy (both 3 → 1 and 2 → 1 end in state 1) → λ′.

CED 15.3.A.3 · Read this in Fix

1. A1.00
   A student who thinks the binding energy is the same in every state picks this. An atom in the n = 3 state is only E₀/9 below the ionized state, so much less energy is needed, and the photon's wavelength is longer.
2. B0.11
   A student who thinks a smaller energy goes with a proportionally shorter wavelength picks this, dividing λ₁ by 9. A photon's energy is hc/λ, so the photon with one-ninth of the energy has 9 times the wavelength.
3. C9.00 Correct
   The energy needed to ionize the atom from state n is 0 − E_n = E₀/n²: E₀ from n = 1 and E₀/9 from n = 3. Since λ = hc/E, a photon with one-ninth of the energy has 9 times the wavelength, so λ₃/λ₁ = 9.00.
4. D1.13
   A student who takes the binding energy of the n = 3 state to be the energy needed to raise the atom from n = 1 to n = 3, 8E₀/9, picks this: λ₃/λ₁ = E₀/(8E₀/9) = 9/8 ≈ 1.13. That energy has already been given to the atom; from n = 3 it needs only E₀/9 more to be ionized.

Working Energy needed to ionize from state n: 0 − E_n = E₀/n². n = 1: E₀; n = 3: E₀/9. λ = hc/E, so λ₃/λ₁ = E₀/(E₀/9) = 9.00. Errors: the same binding energy in every state → 1.00; wavelength taken as proportional to energy → 1/9 = 0.11; binding energy of n = 3 taken as E₃ − E₁ = 8E₀/9 → 9/8 = 1.13.

CED 15.3.A.5 · Read this in Fix

1. A9hc/B
   A student who takes B/9 to be the energy that raised the atom from the ground state to the excited state picks this, using B/9 as the photon energy. B/9 is the energy from the excited state up to ionization; the gap between the two states is B − B/9 = 8B/9.
2. B(9/8)hc/B Correct
   Measured from ionization at 0, the ground state is at −B and the excited state at −B/9. The photon carries the difference, 8B/9, so λ = hc/(8B/9) = (9/8)hc/B.
3. C(9/10)hc/B
   A student who adds the magnitudes of the two level energies, B + B/9 = 10B/9, picks this. Both levels are below 0, so their energy difference is B − B/9 = 8B/9, which is less than B.
4. Dhc/B
   A student who takes the photon energy to be the magnitude of the energy of the level the atom ends in, B, picks this. The photon carries the difference between the two levels, 8B/9.

Working The binding energy of a state is the energy needed to take the atom from that state to ionization (0), so the ground-state energy is −B and the excited-state energy is −B/9. Photon energy: ΔE = (−B/9) − (−B) = 8B/9. Wavelength: λ = hc/ΔE = (9/8)hc/B. Errors: the excited state's binding energy taken as its excitation energy → ΔE = B/9 → 9hc/B; magnitudes added → ΔE = 10B/9 → (9/10)hc/B; photon energy = magnitude of the final level's energy → ΔE = B → hc/B.

CED 15.3.A.5 · Read this in Fix

1. A4ke²/(9hr)
   A student who measures the transition energy from the ground state picks this, using E(9r) − E(r) = 4ke²/(9r). The atom ends with the 4r orbit, not in the ground state, so the photon's energy is E(9r) − E(4r).
2. Bke²/(8hr)
   A student who takes the photon energy to be the magnitude of the energy of the final state, ke²/(8r), picks this. The photon's energy is the difference between the energies of the two states.
3. C13ke²/(72hr)
   A student who adds the magnitudes of the two energies, ke²/(18r) + ke²/(8r) = 13ke²/(72r), picks this. Both energies are negative, so their difference is ke²/(8r) − ke²/(18r) = 5ke²/(72r).
4. D5ke²/(72hr) Correct
   The atom's energy is −ke²/(18r) with the larger orbit and −ke²/(8r) with the smaller one. The photon carries the difference, ke²/(8r) − ke²/(18r) = 5ke²/(72r), so f = ΔE/h = 5ke²/(72hr).

Working E(9r) = −ke²/(18r); E(4r) = −ke²/(8r). ΔE = E(9r) − E(4r) = ke²/(8r) − ke²/(18r) = ke²(9 − 4)/(72r) = 5ke²/(72r). f = ΔE/h = 5ke²/(72hr) (about 4.6 × 10¹⁴ Hz with r = 5.3 × 10⁻¹¹ m, the red line of hydrogen). Errors: energy measured from the ground state, E(9r) − E(r) = 4ke²/(9r) → 4ke²/(9hr); photon energy = |E(4r)| = ke²/(8r) → ke²/(8hr); magnitudes added, ke²/(18r) + ke²/(8r) = 13ke²/(72r) → 13ke²/(72hr).

CED 15.3.A.2.ii · Read this in Fix

1. A178 nm
   A student who thinks a larger energy difference gives a longer wavelength picks the largest gap, 7.00 eV (−1.00 eV → −8.00 eV), which gives 178 nm. Since λ = hc/ΔE, that transition gives the shortest wavelength, not the longest.
2. B414 nm
   A student who thinks only transitions that end in the ground state emit light picks this, from the smallest such gap, 3.00 eV (−5.00 eV → −8.00 eV). Atoms in the upper levels can also drop to lower excited levels, and the smallest gap of all, 1.50 eV, gives a longer wavelength.
3. C355 nm
   A student who takes the photon energy to be the magnitude of the energy of the level the atom ends in picks this, using the highest possible final level, 3.50 eV. The photon's energy is the difference between two levels, and the smallest difference is 1.50 eV.
4. D829 nm Correct
   A longer wavelength means a smaller photon energy, so the longest wavelength comes from the smallest energy difference, 1.50 eV, between −3.50 eV and −5.00 eV: λ = hc/E = 829 nm.

Working The longest wavelength comes from the smallest energy difference between two levels. The gaps between neighboring levels are 3.00 eV, 1.50 eV, and 2.50 eV, so the smallest is 1.50 eV (−3.50 eV → −5.00 eV). E = (1.50 eV)(1.60 × 10⁻¹⁹ J/eV) = 2.40 × 10⁻¹⁹ J. λ = hc/E = (6.63 × 10⁻³⁴ J·s)(3.00 × 10⁸ m/s)/(2.40 × 10⁻¹⁹ J) = 8.29 × 10⁻⁷ m = 829 nm. Errors: the largest gap (7.00 eV) taken to give the longest wavelength → 178 nm; only transitions to the ground state counted (smallest 3.00 eV) → 414 nm; photon energy = magnitude of the final level's energy (smallest possible, 3.50 eV) → 355 nm.

CED 15.3.A.2.ii · Read this in Fix