2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
The energy of level n = 2 of a hydrogen atom is −5.45 × 10⁻¹⁹ J. A hydrogen atom in level n = 2 absorbs a photon of wavelength 486 nm. What is the energy of the atom after it absorbs the photon?
Answer and reasoning
A−1.36 × 10⁻¹⁹ JCorrect The photon's energy is hc/λ = 4.09 × 10⁻¹⁹ J, and absorbing it raises the atom's energy by exactly that amount: −5.45 × 10⁻¹⁹ J + 4.09 × 10⁻¹⁹ J = −1.36 × 10⁻¹⁹ J. The new energy is less negative, so it is higher; it is the energy of level n = 4.
B−9.54 × 10⁻¹⁹ J A student who thinks absorbing a photon lowers an atom's energy subtracts the photon's energy: −5.45 × 10⁻¹⁹ J − 4.09 × 10⁻¹⁹ J = −9.54 × 10⁻¹⁹ J. Absorption adds the photon's energy to the atom, so its energy rises toward zero.
C+9.54 × 10⁻¹⁹ J A student who treats the negative sign of the level energy as a label adds the photon's energy to +5.45 × 10⁻¹⁹ J. The level energy really is negative; adding 4.09 × 10⁻¹⁹ J to −5.45 × 10⁻¹⁹ J gives −1.36 × 10⁻¹⁹ J.
D+4.09 × 10⁻¹⁹ J A student who thinks the atom's new energy equals the energy of the photon it absorbed picks this. The photon's energy, 4.09 × 10⁻¹⁹ J, is the change in the atom's energy, so it must be added to the starting energy, −5.45 × 10⁻¹⁹ J.
Working Photon energy: E = hc/λ = (6.626 × 10⁻³⁴ J s)(2.998 × 10⁸ m s⁻¹)/(486 × 10⁻⁹ m) = 4.09 × 10⁻¹⁹ J. Absorption raises the atom's energy by this amount: −5.45 × 10⁻¹⁹ J + 4.09 × 10⁻¹⁹ J = −1.36 × 10⁻¹⁹ J, the energy of level n = 4. Distractors: subtracting the photon's energy gives −9.54 × 10⁻¹⁹ J; adding it to +5.45 × 10⁻¹⁹ J gives +9.54 × 10⁻¹⁹ J; the photon's energy alone is +4.09 × 10⁻¹⁹ J.
When a molecule absorbs one photon, the energy of the molecule increases by ΔE. Which expression gives the wavelength, λ, of the absorbed photon, where h is Planck's constant and c is the speed of light?
Answer and reasoning
Aλ = ΔE/(hc) A student who thinks photon energy increases with wavelength writes λ in proportion to ΔE. Since ΔE = hc/λ, the wavelength is inversely proportional to ΔE, not proportional to it.
Bλ = h/ΔE A student who takes the frequency to be 1/λ writes ΔE = h/λ and so λ = h/ΔE. The frequency is c/λ, so the speed of light must appear: λ = hc/ΔE.
Cλ = c·ΔE/h A student who thinks wavelength increases with frequency writes λ = cν and, with ν = ΔE/h, gets λ = c·ΔE/h. From c = λν, λ = c/ν = hc/ΔE.
Dλ = hc/ΔECorrect The photon's energy equals the increase in the molecule's energy, ΔE. Combining E = hν with c = λν gives ΔE = hc/λ, so λ = hc/ΔE: the larger the energy change, the shorter the wavelength.
Working The photon's energy equals ΔE. E = hν and ν = c/λ, so ΔE = hc/λ and λ = hc/ΔE.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
3.12.A.1 Energy level Fix
Energy level
One of the allowed energies of an atom or molecule. In energy-level diagrams of atoms, the energies are often measured from zero for the electron removed from the atom, so the levels of the atom have negative values and a less negative value is a higher energy.
Electronic transition
A change of an atom or molecule from one energy level to another, shown on an energy-level diagram as an arrow pointing up (energy gained) or down (energy lost).
Absorption of a photon
When an atom or molecule absorbs a photon, the photon disappears and the energy of the species increases by an amount equal to the photon's energy. For a transition between two levels, the photon's energy equals the difference between their energies.
Emission of a photon
When an atom or molecule moves to a lower energy level it can emit a photon; the energy of the species decreases by an amount equal to the photon's energy, which equals the difference between the two levels.
Line spectrum
The set of separate wavelengths emitted (or absorbed) by the atoms of an element. Each line corresponds to photons of one energy, equal to one energy difference between two levels of the atom.
Students often think The energy of a photon absorbed or emitted in a transition equals the energy of one of the levels involved (the level the atom starts in or ends in), rather than the difference between the two levels. In fact No. The photon's energy equals the difference between the energies of the two levels involved. It equals the energy of a single level only in the special case where the other level is assigned zero energy.
Students often think Absorption and emission work the other way round: an atom gives out a photon when it moves to a higher level, and absorbing a photon lowers its energy. In fact Up. When an atom absorbs a photon its energy increases by the photon's energy and it moves to a higher level; when it emits a photon its energy decreases by the photon's energy and it moves to a lower level.
3.12.A.2 Photon Fix
Photon
A quantum (packet) of electromagnetic radiation. Each photon of radiation of frequency ν carries energy E = hν.
Wavelength (λ)
The distance between successive crests of an electromagnetic wave, usually given in m or nm (1 nm = 10⁻⁹ m).
Frequency (ν)
The number of wave cycles that pass a point each second, in s⁻¹ (hertz, Hz). The symbol is the Greek letter nu, not the v used for speed.
Speed of light and c = λν
In a vacuum all electromagnetic radiation travels at the same speed, c = 2.998 × 10⁸ m s⁻¹. Because c = λν, wavelength and frequency are inversely proportional: radiation with a shorter wavelength has a higher frequency.
Planck's equation
E = hν gives the energy of one photon, where h = 6.626 × 10⁻³⁴ J s is Planck's constant. Combined with c = λν it gives E = hc/λ, so photon energy is proportional to frequency and inversely proportional to wavelength.
Photon energy versus beam power
The energy of one photon depends only on the frequency of the radiation. The power (or brightness) of a beam of one frequency depends on how many photons it delivers per second.
Students often think Photons of longer-wavelength light carry more energy, so photon energy increases with wavelength and red light is more energetic than violet light. In fact No. E = hν and ν = c/λ, so E = hc/λ: the energy of a photon is inversely proportional to its wavelength. A photon of red light (about 700 nm) carries less energy than a photon of violet light (about 400 nm).
Students often think A brighter or more powerful light source emits photons of greater energy, so the energy of each photon increases with the power or brightness of the beam. In fact No. The power or brightness of a beam depends on how many photons it delivers per second as well as on the energy of each photon. The energy of one photon depends only on the frequency of the light, E = hν.
9 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 9
The diagram shows the four lowest energy levels of a hypothetical atom, with each energy measured from level 1, and four transitions labeled P, Q, R, and S. Which arrow represents the transition in which the atom emits the photon of shortest wavelength?
Answer and reasoning
AArrow PCorrect P is a downward transition from 4.2 × 10⁻¹⁹ J to 0, so the atom's energy falls by 4.2 × 10⁻¹⁹ J and it emits a photon of that energy. This is the largest decrease of the three downward arrows (Q, 2.0 × 10⁻¹⁹ J; S, 1.2 × 10⁻¹⁹ J), and since E = hc/λ, the photon with the most energy has the shortest wavelength.
BArrow Q A student who takes the photon's energy to be the energy of the level the atom starts in picks this, because Q starts at the highest level, 5.0 × 10⁻¹⁹ J. The photon's energy is the difference between the two levels, 5.0 − 3.0 = 2.0 × 10⁻¹⁹ J, less than the 4.2 × 10⁻¹⁹ J of transition P.
CArrow R A student who thinks an atom emits a photon when it moves to a higher level picks this, the longest arrow. R points upward: the atom's energy increases, so R represents the absorption of a photon, not an emission.
DArrow S A student who thinks photons of longer wavelength carry more energy reasons that the shortest wavelength belongs to the least energetic photon and picks the smallest downward gap, S (1.2 × 10⁻¹⁹ J). Since E = hc/λ, the shortest wavelength belongs to the photon of greatest energy.
Working An atom emits a photon in a downward transition, and the photon's energy equals the decrease in the atom's energy; by E = hc/λ the most energetic photon has the shortest wavelength. Downward arrows: Q, 5.0 → 3.0 (2.0 × 10⁻¹⁹ J); P, 4.2 → 0 (4.2 × 10⁻¹⁹ J); S, 4.2 → 3.0 (1.2 × 10⁻¹⁹ J). R points upward (absorption). The largest decrease is P.
A hypothetical atom has three lowest energy levels, E₁, E₂, and E₃, in order of increasing energy. A student heats a sample of the gaseous element so that some atoms are raised to E₂ and some to E₃; no atoms reach any higher level. Counting every possible change between two of these three levels, how many of the changes result in the atom emitting a photon?
Answer and reasoning
A1 A student who thinks each element emits light of a single wavelength expects a single emitting transition. The atoms can make three different downward changes, E₃ → E₂, E₂ → E₁, and E₃ → E₁, and each one emits a photon.
B3Correct An atom emits a photon only when it moves to a lower level: E₃ → E₂, E₂ → E₁, and E₃ → E₁. The three upward changes require the atom to gain energy.
C2 A student who thinks an excited atom always returns directly to E₁ counts only E₃ → E₁ and E₂ → E₁. An atom at E₃ can also drop to E₂, emitting a photon, before it drops to E₁, so E₃ → E₂ is a third change that emits a photon.
D6 A student who thinks an atom gives off light whenever it changes level counts all six changes, three upward and three downward. Only the three downward changes emit photons.
Working There are six possible changes between two of the three levels: three upward (E₁ → E₂, E₁ → E₃, E₂ → E₃) and three downward (E₃ → E₂, E₂ → E₁, E₃ → E₁). An atom emits a photon only when its energy decreases, so only the three downward changes emit photons: 3.
One photon of red light, of wavelength 700 nm, and one photon of violet light, of wavelength 400 nm, travel through a vacuum. Which photon has more energy, and why?
Answer and reasoning
AThe red photon, because red light has the longer wavelength A student who thinks longer-wavelength light carries more energy picks this. E = hc/λ, so photon energy is inversely proportional to wavelength: the red photon, with the longer wavelength, has less energy.
BThe violet photon, because violet light travels faster in a vacuum A student who thinks higher-frequency light travels faster picks this. The violet photon does have more energy, but in a vacuum red and violet light travel at the same speed, c; the violet photon has more energy because of its higher frequency.
CThe red photon, because red light has the higher frequency A student who thinks frequency increases with wavelength picks this, giving red light the higher frequency. Since c = λν with c fixed, the longer wavelength of red light means a lower frequency, so the red photon has less energy.
DThe violet photon, because violet light has the higher frequencyCorrect In a vacuum all light travels at c, so ν = c/λ: violet light, with the shorter wavelength, has the higher frequency (7.50 × 10¹⁴ s⁻¹, compared with 4.28 × 10¹⁴ s⁻¹ for the red light). By E = hν, the violet photon carries more energy.
A green laser pointer emits light of wavelength 532 nm. What is the energy of one photon of this light?
Answer and reasoning
A1.25 × 10⁻²⁷ J A student who takes the frequency to be simply 1/λ calculates E = h/λ = (6.626 × 10⁻³⁴ J s)/(5.32 × 10⁻⁷ m) = 1.25 × 10⁻²⁷. The frequency is c/λ; leaving out c gives a result in J s m⁻¹, not J, and a value far too small.
B1.99 × 10⁻²⁵ J A student who reads the ν in E = hν as the speed of light calculates E = hc = 1.99 × 10⁻²⁵ for every photon, whatever its wavelength. ν is the frequency, c/λ = 5.64 × 10¹⁴ s⁻¹, so E = hν = 3.73 × 10⁻¹⁹ J.
C3.73 × 10⁻¹⁹ JCorrect ν = c/λ = (2.998 × 10⁸ m s⁻¹)/(5.32 × 10⁻⁷ m) = 5.64 × 10¹⁴ s⁻¹, and E = hν = (6.626 × 10⁻³⁴ J s)(5.64 × 10¹⁴ s⁻¹) = 3.73 × 10⁻¹⁹ J, the energy of one photon.
D6.20 × 10⁻⁴³ J A student who thinks Planck's equation gives the energy of a mole of photons divides hc/λ by Avogadro's number. hc/λ = 3.73 × 10⁻¹⁹ J is already the energy of one photon; multiplying it by Avogadro's number would give the energy of a mole of photons.
Working ν = c/λ = (2.998 × 10⁸ m s⁻¹)/(532 × 10⁻⁹ m) = 5.64 × 10¹⁴ s⁻¹. E = hν = (6.626 × 10⁻³⁴ J s)(5.64 × 10¹⁴ s⁻¹) = 3.73 × 10⁻¹⁹ J. Distractors: ν = 1/λ gives E = h/λ = 1.25 × 10⁻²⁷; ν read as the speed of light gives E = hc = 1.99 × 10⁻²⁵; hc/λ divided by Avogadro's number gives 6.20 × 10⁻⁴³.
The graph shows the visible emission spectrum of a hypothetical element. Each line is produced when atoms of the element move from a higher energy level to a lower one. Based on the graph, what is the greatest decrease in the energy of one atom that produces a line in this spectrum?
Answer and reasoning
A3.20 × 10⁻¹⁹ J A student who thinks longer-wavelength light carries more energy per photon uses the 620 nm line: hc/λ = 3.20 × 10⁻¹⁹ J. Photon energy is inversely proportional to wavelength, so the 620 nm line comes from the smallest energy decrease, not the greatest.
B4.90 × 10⁻¹⁹ JCorrect Each photon carries away the energy lost by the atom, E = hc/λ, so the greatest decrease gives the line of shortest wavelength, 405 nm: E = (6.626 × 10⁻³⁴ J s)(2.998 × 10⁸ m s⁻¹)/(4.05 × 10⁻⁷ m) = 4.90 × 10⁻¹⁹ J.
C3.64 × 10⁻¹⁹ J A student who thinks brighter light is made of more energetic photons uses the tallest line, 545 nm: hc/λ = 3.64 × 10⁻¹⁹ J. The height of a line shows how many photons of that wavelength are emitted, not how much energy each carries.
D1.64 × 10⁻²⁷ J A student who takes the frequency to be simply 1/λ calculates h/λ for the 405 nm line, 1.64 × 10⁻²⁷. The frequency is c/λ, so E = hc/λ = 4.90 × 10⁻¹⁹ J; without c the result is not even in joules.
Working The decrease in the atom's energy equals the energy of the photon emitted, E = hc/λ, so the greatest decrease produces the shortest-wavelength line, 405 nm: E = (6.626 × 10⁻³⁴ J s)(2.998 × 10⁸ m s⁻¹)/(405 × 10⁻⁹ m) = 4.90 × 10⁻¹⁹ J. Distractors: the longest-wavelength line, 620 nm, gives 3.20 × 10⁻¹⁹ J; the most intense line, 545 nm, gives 3.64 × 10⁻¹⁹ J; E = h/λ at 405 nm gives 1.64 × 10⁻²⁷.
Laser X emits light of wavelength 400 nm with a power of 1.00 mW. Laser Y emits light of wavelength 800 nm with a power of 4.00 mW. What is the ratio of the energy of one photon from laser X to the energy of one photon from laser Y?
Answer and reasoning
A0.50 A student who thinks photon energy increases with wavelength takes EX/EY = 400/800 = 0.50. Since E = hc/λ, the shorter-wavelength light from laser X has the more energetic photons.
B0.25 A student who thinks a more powerful beam is made of more energetic photons takes the ratio of the powers, 1.00/4.00 = 0.25. Power depends on how many photons are emitted per second as well as on the energy of each; the energy of one photon depends only on its frequency.
C2.00Correct The energy of one photon is E = hc/λ, inversely proportional to the wavelength, so EX/EY = λY/λX = 800/400 = 2.00. Laser Y's greater power means that it emits more photons per second, not more energetic photons.
D1.00 A student who reads the ν in E = hν as the speed of light gives every photon the energy hc, so the ratio is 1.00. ν is the frequency, c/λ, which is twice as great for laser X as for laser Y.
Working E = hc/λ, so EX/EY = λY/λX = (800 nm)/(400 nm) = 2.00. The powers do not affect the energy of one photon; they set how many photons each laser emits per second. Distractors: λX/λY = 0.50; ratio of powers 1.00/4.00 = 0.25; E = hc for every photon gives 1.00.
The diagram shows four energy levels of a hypothetical atom and one transition of the atom, shown by the arrow. Which of the following describes the transition?
Answer and reasoning
AThe atom emits a photon of energy 7.5 × 10⁻¹⁹ J, and the atom's energy decreases.Correct The arrow runs from −4.5 × 10⁻¹⁹ J down to −12.0 × 10⁻¹⁹ J. The atom's energy decreases by the gap between the two levels, 7.5 × 10⁻¹⁹ J, and that energy leaves as an emitted photon.
BThe atom absorbs a photon of energy 7.5 × 10⁻¹⁹ J, and the atom's energy decreases. A student who has absorption and emission the wrong way round picks this, taking a move to a lower level as absorption. Absorbing a photon raises an atom's energy; a downward arrow is the emission of a photon.
CThe atom absorbs a photon of energy 7.5 × 10⁻¹⁹ J, and the atom's energy rises. A student who ignores the negative signs picks this, reading the change as 4.5 to 12.0, a gain of 7.5 × 10⁻¹⁹ J. Because the values are negative, −12.0 × 10⁻¹⁹ J is the lower energy, so the atom loses energy and emits the photon.
DThe atom emits a photon of energy 12.0 × 10⁻¹⁹ J, and the atom's energy decreases. A student who matches the photon to the labeled value of the level the atom ends in picks this. The photon's energy is the difference between the two levels, 12.0 − 4.5 = 7.5 in units of 10⁻¹⁹ J.
Working The arrow points from Level 2 (−4.5 × 10⁻¹⁹ J) down to Level 1 (−12.0 × 10⁻¹⁹ J), so the atom's energy decreases by (−4.5 × 10⁻¹⁹ J) − (−12.0 × 10⁻¹⁹ J) = 7.5 × 10⁻¹⁹ J. A decrease in the atom's energy means a photon of that energy is emitted.
A sodium lamp emits light of wavelength 589 nm. How many photons of this light carry a total energy of 2.00 J?
Answer and reasoning
A1.78 × 10²⁷ A student who takes the frequency as 1/λ picks this, using E = h/λ = 1.12 × 10⁻²⁷ for one photon. The frequency is ν = c/λ, so the energy of one photon is hc/λ = 3.37 × 10⁻¹⁹ J.
B5.93 × 10¹⁸Correct One photon has E = hc/λ = (6.626 × 10⁻³⁴ J·s)(2.998 × 10⁸ m/s) ÷ (5.89 × 10⁻⁷ m) = 3.37 × 10⁻¹⁹ J. Dividing the total energy by the energy of one photon gives 2.00 J ÷ 3.37 × 10⁻¹⁹ J = 5.93 × 10¹⁸ photons.
C1.01 × 10²⁵ A student who reads ν in E = hν as the speed of light picks this, using h × c = 1.99 × 10⁻²⁵ for one photon. ν is the frequency, c/λ = 5.09 × 10¹⁴ s⁻¹, so one photon has 3.37 × 10⁻¹⁹ J.
D3.57 × 10⁴² A student who thinks hc/λ is the energy of a mole of photons picks this, dividing 3.37 × 10⁻¹⁹ J by Avogadro's number before dividing it into 2.00 J. Planck's equation already gives the energy of one photon.
Working Energy of one photon: E = hν = hc/λ = (6.626 × 10⁻³⁴ J·s)(2.998 × 10⁸ m/s) ÷ (589 × 10⁻⁹ m) = 3.37 × 10⁻¹⁹ J. Number of photons = 2.00 J ÷ 3.37 × 10⁻¹⁹ J = 5.93 × 10¹⁸.
A gaseous atom moves from a higher energy level to a lower energy level and emits one photon. A student wants to use Planck's constant to calculate the frequency of the photon. Which additional information does the student need?
Answer and reasoning
AThe energy of the level in which the atom starts A student who thinks the photon's energy equals the energy of one of the levels picks this. The photon carries away only the difference between the two levels, so the energy of the starting level alone does not fix its frequency.
BThe brightness of the light that the gas gives off A student who thinks brighter light is made of more energetic photons picks this. Brightness depends on how many photons are emitted each second; the energy and frequency of each photon are fixed by the gap between the two levels.
CThe difference in energy between the two levelsCorrect The energy of the atom decreases by the energy of the photon, so the photon's energy is the gap between the two levels. Dividing that gap by Planck's constant gives the frequency, ν = E/h.
DThe speed at which the emitted photon travels A student who thinks light of higher frequency travels faster picks this, expecting the speed to reveal the frequency. Every photon travels at the same speed, c, in a vacuum, so the speed says nothing about the frequency of this photon.
Working The atom's energy decreases by an amount equal to the energy of the photon, so Ephoton = Ehigher − Elower. Then Ephoton = hν gives ν = (Ehigher − Elower)/h. With h known, the quantity needed is the difference between the energies of the two levels.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
