Calculate the energy of a photon of wavelength 450 nm in joules and electronvolts. Use h = 6.63 x 10−34 J s, c = 3.00 x 108 m/s and 1 eV = 1.60 x 10−19 J.
[4 marks]
Total for this question: 4
All equation deep-dives
Photon energy and de Broglie wavelength
Particles and radiation
AQA 7408
In the data booklet
ORIGINAL
E = h f and lambda = h / p
Derived rearrangement
E = h c / lambdaDerived from the given E = h f and c = f lambda; it is not printed verbatim.Derived rearrangement
lambda = h c / EA rearrangement of the derived E = h c / lambda form.Nanometres and electronvolts are the traps: convert to metres and joules before using the booklet equations.
Board and spec code confirmed against AQA 7408 · registry checked 2026-07-11How this checking works
Know the equation
| Symbol | Quantity | Unit |
|---|---|---|
| E | photon energy | J |
| h | Planck constant | J s |
| f | photon frequency | Hz |
| c | speed of light in vacuum | m/s |
| lambda | wavelength | m |
| p | particle momentum | kg m/s |
| m | particle mass | kg |
| v | particle speed | m/s |
Rearrangements
- f = E / h
- p = h / lambda
- lambda = h / (m v)
Apply it — mark your own working
Work each one out on paper first, then reveal the mark scheme and tick the marks you actually earned. That is exactly how you should mark past papers.
Q1
Do the calculation on paper first — then mark it.
Q2
An electron of mass 9.11 x 10−31 kg moves at 2.5 x 106 m/s. Calculate its de Broglie wavelength. Use h = 6.63 x 10−34 J s.
[3 marks]
Total for this question: 3
Do the calculation on paper first — then mark it.
Where the marks get lost
- Using wavelength in nanometres without multiplying by 10−9.
- Converting joules to electronvolts in the wrong direction; divide the joule value by 1.60 x 10−19.
- Using lambda = h/m and forgetting the velocity in momentum p = m v.
Exam tip: Write the momentum step explicitly before using lambda = h/p. It earns the method clearly and prevents a missing mass or speed factor.
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