Properties of Photons: Study Guide (Chemistry 3.12)

A photon is a fixed packet of energy

In Topic 3.11 you saw that atoms and molecules have quantized energy levels, and that each region of the spectrum matches one kind of transition. This topic puts numbers on that idea.

Light arrives in photons. Each photon carries one fixed amount of energy that depends only on its frequency. When a photon meets an atom or molecule, there are two possibilities:

• Absorption. The photon disappears, and the energy of the atom or molecule goes up by exactly the energy of the photon.
• Emission. The atom or molecule drops to a lower level, and its energy goes down by exactly the energy of the photon it gives out.

Energy is conserved in both cases, so:

E(photon) = ΔE(atom or molecule) = E(upper level) − E(lower level)

Because the levels are quantized, only photons with energies equal to a gap can be absorbed or emitted. That is why the spectrum of an element shows sharp lines at a few wavelengths, not a continuous smear of colour.

The two equations

Wavelength and frequency. Every electromagnetic wave travels through a vacuum at the same speed, the speed of light c. Speed = distance per wave × waves per second, so:

c = λν

• c = 2.998 × 10⁸ m s⁻¹ (speed of light)
• λ (lambda) = wavelength, in metres
• ν (nu) = frequency, in s⁻¹ (hertz, Hz)

Since c is fixed, λ and ν are inversely proportional. Halve the wavelength and the frequency doubles.

Energy and frequency (Planck’s equation).

E = hν

• E = energy of one photon, in joules
• h = 6.626 × 10⁻³⁴ J s (Planck’s constant)

Energy is directly proportional to frequency. Double the frequency and each photon carries twice the energy.

Combining them. Substitute ν = c/λ into E = hν:

E = hc / λ

This lets you go straight from wavelength to energy. Energy is inversely proportional to wavelength, which is the “shorter wavelength, more energy” rule from Topic 3.11 written as an equation.

Quantity	Symbol	Unit you must use	Typical size for visible light
Wavelength	λ	m	about 4 × 10⁻⁷ to 7 × 10⁻⁷ m
Frequency	ν	s⁻¹ (Hz)	about 4 × 10¹⁴ to 7.5 × 10¹⁴ s⁻¹
Energy per photon	E	J	about 3 × 10⁻¹⁹ to 5 × 10⁻¹⁹ J
Energy per mole of photons	E × N_A	J mol⁻¹, then kJ mol⁻¹	about 170 to 300 kJ mol⁻¹

Per photon or per mole?

E = hν gives the energy of one photon, so the number is tiny (around 10⁻¹⁹ J). Chemists usually compare energies per mole, for example with bond energies in kJ mol⁻¹. To convert, multiply by Avogadro’s number and divide by 1000:

energy per mole (kJ mol⁻¹) = E(per photon, J) × 6.022 × 10²³ mol⁻¹ ÷ 1000 J kJ⁻¹

The reverse also works: to find the energy one molecule needs to break a bond, divide the bond energy in J mol⁻¹ by N_A.

Choosing a route

Most photon questions are one of four routes. Decide which one before you touch the calculator.

You are given	You want	Route
wavelength λ	energy per photon	convert λ to m, then E = hc/λ
frequency ν	wavelength	λ = c/ν, then convert m to nm if asked
energy per mole (kJ mol⁻¹)	wavelength	× 1000, ÷ N_A to get J per photon, then λ = hc/E
two energy levels	wavelength of the line	ΔE = upper − lower, then λ = hc/ΔE

Writing each step as a fraction lets the units check you, just as in mole calculations. For example, to find the wavelength that matches 250 kJ mol⁻¹:

(250 kJ mol⁻¹) × (1000 J / 1 kJ) × (1 mol / 6.022 × 10²³ photons) = 4.151 × 10⁻¹⁹ J per photon, then λ = hc / E = 4.785 × 10⁻⁷ m = 479 nm.

If the units do not cancel to the one you want, the setup is wrong.

Worked example 1: one photon and one mole of photons

Question. A particular blue LED emits light of wavelength 455 nm. Calculate (a) the frequency, (b) the energy of one photon and (c) the energy of one mole of these photons, in kJ mol⁻¹.

1. Convert the wavelength to metres. 455 nm × (10⁻⁹ m / 1 nm) = 4.55 × 10⁻⁷ m.
2. (a) Frequency. ν = c / λ = (2.998 × 10⁸ m s⁻¹) ÷ (4.55 × 10⁻⁷ m) = 6.589 × 10¹⁴ s⁻¹.
3. (b) Energy of one photon. E = hν = (6.626 × 10⁻³⁴ J s) × (6.589 × 10¹⁴ s⁻¹) = 4.366 × 10⁻¹⁹ J. The units cancel: J s × s⁻¹ = J.
4. (c) Energy per mole. 4.366 × 10⁻¹⁹ J × 6.022 × 10²³ mol⁻¹ = 2.629 × 10⁵ J mol⁻¹ = 263 kJ mol⁻¹.

Answer. (a) 6.59 × 10¹⁴ s⁻¹; (b) 4.37 × 10⁻¹⁹ J; (c) 263 kJ mol⁻¹ (3 significant figures, matching 455 nm).

Check. Using E = hc/λ in one step: (6.626 × 10⁻³⁴ × 2.998 × 10⁸) ÷ 4.55 × 10⁻⁷ = 4.366 × 10⁻¹⁹ J, the same. The frequency and energy sit inside the visible-light ranges in the table above, which is right for blue light.

Reading an energy-level diagram

An energy-level diagram draws each allowed level as a horizontal line, with energy increasing upwards. An arrow down is emission; an arrow up is absorption. The length of the arrow is the photon’s energy. Levels are often given negative energies, measured from the point where the electron would be removed completely (energy zero). Only differences between levels matter for photons, so the negative signs cancel out when you subtract.

Worked example 2: emission lines from an energy-level diagram

Question. The fictional atom Z has the three energy levels shown in Figure 1. Excited Z atoms can drop from level 3 to level 2, from level 2 to level 1, or straight from level 3 to level 1. Calculate the wavelength of each emitted photon and state the region of the spectrum.

1. Find each energy gap (upper minus lower):
   • 3 → 2: (−2.10 × 10⁻¹⁹ J) − (−5.20 × 10⁻¹⁹ J) = 3.10 × 10⁻¹⁹ J
   • 2 → 1: (−5.20 × 10⁻¹⁹ J) − (−9.00 × 10⁻¹⁹ J) = 3.80 × 10⁻¹⁹ J
   • 3 → 1: (−2.10 × 10⁻¹⁹ J) − (−9.00 × 10⁻¹⁹ J) = 6.90 × 10⁻¹⁹ J
2. Use λ = hc / E for each. First, hc = (6.626 × 10⁻³⁴ J s)(2.998 × 10⁸ m s⁻¹) = 1.9865 × 10⁻²⁵ J m.
   • 3 → 2: λ = 1.9865 × 10⁻²⁵ J m ÷ 3.10 × 10⁻¹⁹ J = 6.408 × 10⁻⁷ m = 641 nm, visible (red-orange)
   • 2 → 1: λ = 1.9865 × 10⁻²⁵ ÷ 3.80 × 10⁻¹⁹ = 5.228 × 10⁻⁷ m = 523 nm, visible (green)
   • 3 → 1: λ = 1.9865 × 10⁻²⁵ ÷ 6.90 × 10⁻¹⁹ = 2.879 × 10⁻⁷ m = 288 nm, ultraviolet

Check 1: energy adds up. The 3 → 1 drop equals the two smaller drops together: 3.10 × 10⁻¹⁹ + 3.80 × 10⁻¹⁹ = 6.90 × 10⁻¹⁹ J. An atom that falls in two steps emits two lower-energy photons instead of one high-energy photon.

Check 2: absorption. A ground-state Z atom can absorb 523 nm or 288 nm photons, raising it to level 2 or level 3. It cannot absorb 641 nm light from the ground state, because no gap from level 1 equals 3.10 × 10⁻¹⁹ J.

Note. Only the 3 → 1 line falls outside the visible range, so a spectroscope pointed at glowing Z would show two coloured lines. You would need a UV detector to see the third.

Brightness is not photon energy

A brighter light of the same colour delivers more photons per second, not more energetic photons. Each photon still carries E = hν. So:

• A dim 523 nm lamp can raise ground-state Z atoms (Worked example 2) to level 2, because each of its photons carries exactly the 3.80 × 10⁻¹⁹ J gap.
• A very bright 641 nm lamp cannot, because each of its photons carries only 3.10 × 10⁻¹⁹ J, which matches no gap from level 1. An atom absorbs one photon at a time, so piling on more photons does not help.

Brightness changes how many atoms or molecules undergo a transition each second, not which transition happens. This idea returns in Topic 3.13, where more absorbing particles remove more photons from a beam.

Common misconceptions

• Forgetting to convert nm to m. Using λ = 455 instead of 4.55 × 10⁻⁷ m gives an answer 10⁹ times too small for E. Convert before you substitute.
• Mixing per photon and per mole. E = hν gives joules per photon. Multiply by N_A for J mol⁻¹, then divide by 1000 for kJ mol⁻¹.
• “Longer wavelength means more energy.” E = hc/λ: energy falls as wavelength rises.
• “Wavelength and frequency are directly proportional.” They are inversely proportional; their product is always c.
• “An emitted photon’s energy equals the energy of the level it came from.” It equals the difference between the two levels.
• “An atom can absorb part of a photon’s energy.” A photon is absorbed whole, and only if its energy matches a gap; otherwise it passes through.
• “Brighter light gives more energetic photons.” Brightness is the number of photons; energy per photon depends only on frequency.
• Dropping the negative signs carelessly. With negative level energies, always do upper − lower. The answer for a photon must be positive.

Where this leads

This topic gives numbers to the transitions in Topic 3.11, Spectroscopy and the Electromagnetic Spectrum. Next, Topic 3.13, the Beer-Lambert Law, uses the absorption of photons by a solution to measure its concentration. Try the practice questions now, then use the revision notes and the checklist to consolidate.