AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/3/3-12)
Unit 3 · Topic 3.12
3.12 Properties of Photons
Light comes in packets of energy called photons. Wavelength and frequency are linked by c = λν, and a photon's energy is E = hν, so shorter wavelengths carry more energy. When an atom or molecule absorbs or emits a photon, its energy changes by exactly the photon's energy.
Key terms
- photon
- wavelength (λ)
- frequency (ν)
- Planck's constant (h)
- speed of light (c)
- absorption and emission
Waves: wavelength and frequency
Wavelength (λ, lambda) is the distance from one wave crest to the next, often given in nanometers (1 nm = 10⁻⁹ m). Frequency (ν, nu) is how many waves pass a point per second, in hertz (1 Hz = 1 s⁻¹).
All electromagnetic radiation travels at the speed of light, c = 2.998 × 10⁸ m/s. They're linked by c = λν. Because c is fixed, wavelength and frequency are inversely proportional: when one doubles, the other halves.
Photons: E = hν
Light is absorbed and emitted in packets called photons. The energy of one photon is E = hν, where h is Planck's constant, 6.626 × 10⁻³⁴ J·s. Higher frequency means higher photon energy.
Combining the two equations gives E = hc/λ, which shows that shorter wavelength means higher energy. A violet photon carries more energy than a red one.
Energy levels and transitions
When an atom or molecule absorbs a photon, its energy rises by exactly the photon's energy. When it emits a photon, its energy falls by exactly that much. So ΔE (the gap between two levels) = hν of the photon.
Because energy levels are fixed, only photons whose energy matches a gap can be absorbed. That's why each element has its own set of spectral lines, like a fingerprint. Hydrogen's visible lines include red light at 656 nm and blue-green light at 486 nm.
Reading an energy-level diagram
An energy-level diagram shows allowed energies as horizontal lines, with higher energy at the top. An arrow pointing up is absorption; an arrow pointing down is emission. The length of the arrow represents ΔE.
A longer arrow means a larger energy change, a higher-frequency photon and a shorter wavelength. So if one emission arrow is twice as long as another, its photon has twice the energy and half the wavelength.
Per photon or per mole
E = hν gives the energy of one photon, in joules. To get energy per mole of photons, multiply by Avogadro's number, then convert J to kJ if needed. Reaction energies are usually in kJ/mol, so this conversion lets you compare a photon's energy with, say, a bond energy.
Unit checklist
- Wavelength must be in meters when you use c = 2.998 × 10⁸ m/s. Convert nm by multiplying by 10⁻⁹.
- Frequency is in s⁻¹ (Hz).
- Energy from E = hν is in joules per photon.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
From wavelength to frequency and energy
Hydrogen emits red light with a wavelength of 656 nm. Calculate the frequency, the energy of one photon and the energy of one mole of these photons.
Show the solutionHide the solution
- Step 1: Convert wavelength: 656 nm = 656 × 10⁻⁹ m = 6.56 × 10⁻⁷ m.
- Step 2: Frequency: ν = c/λ = (2.998 × 10⁸ m/s) ÷ (6.56 × 10⁻⁷ m) = 4.57 × 10¹⁴ s⁻¹.
- Step 3: Energy per photon: E = hν = (6.626 × 10⁻³⁴ J·s)(4.57 × 10¹⁴ s⁻¹) = 3.03 × 10⁻¹⁹ J.
- Step 4: Per mole: (3.03 × 10⁻¹⁹ J)(6.022 × 10²³ mol⁻¹) = 1.82 × 10⁵ J/mol = 182 kJ/mol.
Answer: ν = 4.57 × 10¹⁴ s⁻¹; E = 3.03 × 10⁻¹⁹ J per photon; 182 kJ/mol
- Example 2Calculator allowed
From an energy gap to a wavelength
An electron in an atom drops between two levels that differ by 4.09 × 10⁻¹⁹ J. What wavelength of light is emitted, in nm?
Show the solutionHide the solution
- Step 1: The photon carries away exactly ΔE = 4.09 × 10⁻¹⁹ J.
- Step 2: λ = hc/E = (6.626 × 10⁻³⁴ J·s)(2.998 × 10⁸ m/s) ÷ (4.09 × 10⁻¹⁹ J) = 4.86 × 10⁻⁷ m.
- Step 3: Convert: 4.86 × 10⁻⁷ m = 486 nm, which is blue-green visible light.
Answer: 486 nm
- Example 3Calculator allowed
Forgetting to convert nanometers (classic trap)
A student calculates the energy of a 656 nm photon as E = hc/λ = (6.626 × 10⁻³⁴)(2.998 × 10⁸) ÷ 656 and gets 3.03 × 10⁻²⁸ J. What went wrong?
Show the solutionHide the solution
- Step 1: The speed of light is in meters per second, so λ must be in meters.
- Step 2: 656 nm is 6.56 × 10⁻⁷ m, not 656 m.
- Step 3: Using meters gives 3.03 × 10⁻¹⁹ J. The student's answer is off by a factor of 10⁹.
- Step 4: Sense check: visible photons have energies around 10⁻¹⁹ J.
Answer: The wavelength wasn't converted to meters; the correct energy is 3.03 × 10⁻¹⁹ J.
Common mistakes
- Leaving wavelength in nm when using c in m/s.
- Mixing up energy per photon and energy per mole.
- Thinking longer wavelength means higher energy. Energy is inversely proportional to wavelength.
- Confusing the symbols ν (nu, frequency) and v (velocity).
On the exam
- Expect calculations with c = λν and E = hν, both of which are on the equation sheet. Show the conversion of units as a separate step.
- Questions may ask whether a photon has enough energy to cause a particular change, such as breaking a bond. Convert to the same units (J per photon or kJ/mol) before comparing.
Connected topics
Videos
Check yourself
5 questions on 3.12 Properties of Photons. Pick an answer to see if you got it, and why.
What is the energy of one photon of green light with a wavelength of 500. nm? (h = 6.626 × 10⁻³⁴ J·s; c = 3.00 × 10⁸ m/s)
In a hydrogen atom, an electron drops from an energy level of −2.42 × 10⁻¹⁹ J to a level of −5.45 × 10⁻¹⁹ J and emits one photon. What is the wavelength of the photon? (h = 6.626 × 10⁻³⁴ J·s; c = 3.00 × 10⁸ m/s)
| Bond | Bond energy (kJ/mol) |
|---|---|
| Cl–Cl | 242 |
| Br–Br | 193 |
| C–C | 348 |
| H–Cl | 431 |
Average bond energies. Use h = 6.626 × 10⁻³⁴ J·s, c = 3.00 × 10⁸ m/s and 6.022 × 10²³ mol⁻¹.
What is the energy of one mole of photons of violet light with a wavelength of 400. nm?
Which bonds in the table could be broken by absorbing a single photon of 400. nm light?
What is the longest wavelength of light whose photons each carry enough energy to break a Cl–Cl bond?
0 of 5 answered