Photon properties — E = hf, p = h - λ: Worked examples

This page is the practice engine for the parent photon note. There we built the two formulas. Here we drive them through every kind of situation an exam or a real problem can hand you — so that when you see a new question, you already recognise which cell of the table it lives in.

Constants used throughout (memorise these):

The scenario matrix

Before working anything, here is the full landscape. Each row is a class of problem; the last column names the worked example that covers it. If you can do all of these, no photon question can surprise you.

Cell	What varies / the twist	Given → Wanted	Covered by
A. Given $λ$	you know wavelength	$λ \to E$	Example 1
B. Given $f$	you know frequency	$f \to E$	Example 2
C. Momentum route	massless, need $p$	$λ \to p$ , cross-check via $E / c$	Example 3
D. Beam / power	many photons	$P, λ \to N$ per second	Example 4
E. Scaling (ratio)	no numbers, just "×2"	$λ \to E, p$ proportion	Example 5
F. Limiting / degenerate	$λ \to \infty$ (radio), $λ \to 0$ (gamma), $f = 0$	behaviour at extremes	Example 6
G. Push on matter	real world	force from photon stream	Example 7
H. Exam twist	eV given, or work backwards	$E$ (eV) $\to λ$ ; identify colour	Example 8

The figure below is the mental map — every cell is just a different door into the same three formulas.

Figure — Photon properties — E = hf, p = h - λ

Worked Examples

Cell A — you are handed a wavelength

Worked example 1 — Energy of a 400 nm (violet) photon

A photon has wavelength $λ = 400 nm$ . Find its energy in joules and in eV.

Forecast: Violet is high-frequency (short wavelength), so guess a large-ish photon energy — a few eV. Hold that guess.

Convert the wavelength to metres: $λ = 400 \times 1 0^{- 9} m$ . Why this step? Every SI formula wants metres; "nm" would break the units silently.
Pick the form containing $λ$ : $E = \frac{h c}{λ}$ . Why this step? We were given $λ$ , not $f$ — using $h c / λ$ means no extra conversion.
Plug in: $E = \frac{1.986 \times 1 0 ^{- 25}}{400 \times 1 0 ^{- 9}} = 4.97 \times 1 0^{- 19} J$ .
Convert to eV: $E = \frac{4.97 \times 1 0 ^{- 19}}{1.602 \times 1 0 ^{- 19}} = 3.10 eV$ . Why this step? eV is the human-scale unit for photons; $3.10$ eV is instantly comparable to bond/atomic energies.

Verify: Use the shortcut $E (eV) = \frac{1240}{λ ( nm )} = \frac{1240}{400} = 3.10 eV$ ✓. Matches the forecast of "a few eV."

Cell B — you are handed a frequency

Cell C — the momentum route (massless!)

Worked example 3 — Momentum of a 500 nm photon, two ways

Find the momentum of a $λ = 500 nm$ photon, and confirm both routes agree.

Forecast: A photon carries momentum even with zero rest mass. It will be tiny — around $1 0^{- 27} kg\cdotm/s$ . Guess the exponent.

Direct route: $p = \frac{h}{λ} = \frac{6.626 \times 1 0 ^{- 34}}{500 \times 1 0 ^{- 9}} = 1.33 \times 1 0^{- 27} kg\cdotm/s$ . Why this step? $p = h / λ$ needs only the wavelength we were given.
Independent route via energy: from Example 2, $E = 3.98 \times 1 0^{- 19} J$ , and $p = \frac{E}{c}$ . Why this step? For a massless particle the relativistic relation collapses to $E = p c$ , so $p = E / c$ is a separate path — perfect for cross-checking.
Compute: $p = \frac{3.98 \times 1 0 ^{- 19}}{3 \times 1 0 ^{8}} = 1.33 \times 1 0^{- 27} kg\cdotm/s$ .

Verify: Both routes give $1.33 \times 1 0^{- 27} kg\cdotm/s$ ✓. The agreement is the proof that $E = p c$ holds for light. Units: $\frac{J\cdots}{m} = \frac{kg\cdotm ^{2} / s ^{2} \cdot s}{m} = kg\cdotm/s$ ✓. See Relativistic Energy-Momentum Relation for why $E = p c$ .

Cell D — a beam: counting many photons

Worked example 4 — How many photons per second from a 5 mW green pointer (

$λ = 532$ nm)? A laser pointer emits power $P = 5 mW$ at $λ = 532 nm$ . How many photons leave it each second?

Forecast: Each photon is ~ $1 0^{- 19}$ J and the beam pours out $1 0^{- 3}$ J each second, so expect around $1 0^{16}$ photons/s — a mind-bendingly huge number. Guess the power of ten.

Energy per photon: $E = \frac{1240}{532} = 2.33 eV = 2.33 \times 1.602 \times 1 0^{- 19} = 3.73 \times 1 0^{- 19} J$ . Why this step? Power is "energy per second"; to convert it to a photon count we must know the energy of one photon.
Photons per second: $N = \frac{P}{E} = \frac{5 \times 1 0 ^{- 3}}{3.73 \times 1 0 ^{- 19}}$ . Why this step? $P = N \times E$ (count times energy each = total energy per second), so $N = P / E$ .
Compute: $N = 1.34 \times 1 0^{16}$ photons/s.

Verify: Multiply back: $N \times E = 1.34 \times 1 0^{16} \times 3.73 \times 1 0^{- 19} = 5.0 \times 1 0^{- 3} W = 5 mW$ ✓. This is why a "weak" beam still looks perfectly continuous — see Wave-Particle Duality.

Cell E — pure scaling, no numbers needed

Worked example 5 — Triple the wavelength: what happens to

$E$ and $p$ ? A photon's wavelength is increased to 3 times its original value. By what factor do its energy and momentum change?

Forecast: Redder = longer wavelength = weaker photon. Both $E$ and $p$ should shrink. Guess the factor.

Write both formulas as proportions: $E = \frac{h c}{λ} \propto \frac{1}{λ}$ and $p = \frac{h}{λ} \propto \frac{1}{λ}$ . Why this step? When only a ratio is asked, keep the constants symbolic — they cancel.
Replace $λ \to 3 λ$ : new $E = \frac{h c}{3 λ} = \frac{1}{3} E_{old}$ ; same for $p$ . Why this step? Substituting shows the constant $h c$ (or $h$ ) survives untouched, so only the $1/3$ factor matters.

Verify: Both drop to one third. Check with numbers: $λ = 500 \to 1500 nm$ gives $E = 1240/1500 = 0.827 eV$ , and $2.48/3 = 0.827 eV$ ✓. This inverse- $λ$ behaviour is exactly what powers de Broglie Wavelength.

Cell F — the extreme / degenerate cases

Worked example 6 — The three limits: radio, gamma, and "

$f = 0$ " Describe the photon energy for (a) a long radio wave $λ = 3 km$ , (b) a gamma ray $λ = 1 pm = 1 0^{- 12} m$ , and (c) the degenerate case $f \to 0$ .

Forecast: Radio → almost no energy; gamma → enormous energy; $f = 0$ → no photon at all. Confirm each.

(a) Radio, $λ = 3000 m$ : $E = \frac{1240 eV\cdotnm}{3000 \times 1 0 ^{9} nm} = 4.1 \times 1 0^{- 10} eV$ . Why this step? Convert 3000 m to nm ( $\times 1 0^{9}$ ) to reuse the 1240 shortcut. The energy is vanishingly small — one radio photon barely nudges anything, which is why radio behaves so wave-like.
(b) Gamma, $λ = 1 0^{- 12} m = 1 0^{- 3} nm$ : $E = \frac{1240}{1 0 ^{- 3}} = 1.24 \times 1 0^{6} eV = 1.24 MeV$ . Why this step? Same formula, opposite extreme — a single gamma photon carries MeV energies, enough to knock electrons out of atoms (ionising radiation).
(c) $f \to 0$ : $E = h f \to 0$ and $p = E / c \to 0$ . There is no photon — a zero-frequency oscillation carries nothing. Why this step? Checking the degenerate limit confirms the formula behaves sensibly: no oscillation, no quantum.

Verify: As $λ$ sweeps from $3 km$ down to $1 pm$ , $E$ climbs from $\sim 1 0^{- 10} eV$ to $\sim 1 0^{6} eV$ — 15 orders of magnitude, all from one $1/ λ$ law ✓. Figure below plots this. Radio: $1240/ (3000 \times 1 0^{9}) = 4.13 \times 1 0^{- 10}$ eV ✓; Gamma: $1240/1 0^{- 3} = 1.24 \times 1 0^{6}$ eV ✓.

Cell G — the real-world push

Worked example 7 — Force of sunlight fully absorbed on a solar sail

Sunlight of intensity $I = 1360 W/m^{2}$ falls on and is fully absorbed by a sail of area $A = 1000 m^{2}$ . What force does it feel?

Forecast: Light does push (comet tails, solar sails). But with $p = E / c$ and $c$ huge, the force will be small — a few millinewtons. Guess.

Power hitting the sail: $P = I \times A = 1360 \times 1000 = 1.36 \times 1 0^{6} W$ . Why this step? Intensity is power per area; total power is what carries the momentum.
Momentum arriving per second: each joule of light carries $p = E / c$ , so momentum per second $= \frac{P}{c}$ . Why this step? Force is momentum delivered per second (Newton's second law), and photon momentum is $E / c$ — see Radiation Pressure.
Force (full absorption): $F = \frac{P}{c} = \frac{1.36 \times 1 0 ^{6}}{3 \times 1 0 ^{8}} = 4.53 \times 1 0^{- 3} N$ . Why this step? Absorbed light hands over all its momentum once; a mirror would double this (it reflects, reversing momentum).

Verify: $4.53 mN$ on a huge sail — tiny but continuous, and in space with no friction it accelerates a spacecraft over months ✓. Units: $\frac{W}{m/s} = \frac{J/s}{m/s} = \frac{kg\cdotm ^{2} / s ^{3}}{m/s} = kg\cdotm/s^{2} = N$ ✓.

Cell H — the exam twist (work backwards)

Worked example 8 — Given the energy, find the wavelength and name the colour

A photon carries exactly $E = 1.77 eV$ . Find its wavelength and identify the colour.

Forecast: Under 2 eV means low-energy visible light → this should land in the red end, around 700 nm. Guess.

Rearrange the shortcut $E (eV) = \frac{1240}{λ ( nm )}$ for $λ$ : $λ = \frac{1240}{E ( eV )}$ . Why this step? We are given $E$ and want $λ$ — solving the same relation backwards avoids re-deriving from $h c$ .
Plug in: $λ = \frac{1240}{1.77} = 700.6 nm$ .
Identify: 700 nm is deep red (visible band is ~380–700 nm). Why this step? Naming the colour is the sanity check the examiner wants — and it matches Photoelectric Effect intuition that redder = less energetic.

Verify: Forward-check with full constants: $E = \frac{1.986 \times 1 0 ^{- 25}}{700.6 \times 1 0 ^{- 9}} = 2.835 \times 1 0^{- 19} J = 1.77 eV$ ✓. Forecast of "red, ~700 nm" confirmed.

Recall Rapid-fire self-test (cover the answers)

A 400 nm photon's energy in eV? ::: $3.10$ eV A photon of $2.48$ eV — what wavelength? ::: $500$ nm If $λ$ triples, $E$ becomes? ::: one third Gamma ray at 1 pm — rough energy? ::: about $1.24$ MeV Force from fully absorbed power $P$ ? ::: $F = P / c$ Momentum of a 500 nm photon? ::: $1.33 \times 1 0^{- 27}$ kg·m/s

Connections

Photoelectric Effect — Cell H's "colour vs energy" logic is exactly its threshold idea.
Radiation Pressure — the macroscopic form of Cell G.
Relativistic Energy-Momentum Relation — justifies $p = E / c$ used in Cell C.
de Broglie Wavelength — Cell E's $1/ λ$ scaling, run in reverse for matter.
Compton Effect — momentum book-keeping ( $p = h / λ$ ) taken to a collision.
Wave-Particle Duality — why Cell D's huge photon count looks like a smooth wave.
Planck's Law and Blackbody Radiation — where $E = h f$ was born.