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Quantum Physics: Question 1

Syllabus 22.1

Multiple choice A2 1 mark

A helium-neon laser used in a college physics laboratory emits red light of wavelength 632.8 nm632.8\text{ nm}.

What is the energy of one photon of this light, in electron-volts (eV)?

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

Step 1: Calculate the photon energy in joules

The energy of a photon is related to its wavelength by: E=hf=hcλE = hf = \frac{hc}{\lambda}

Substituting h=6.63×10−34 J sh=6.63\times10^{-34}\text{ J s}, c=3.00×108 m s−1c=3.00\times10^{8}\text{ m s}^{-1} and λ=632.8×10−9 m\lambda=632.8\times10^{-9}\text{ m}:

E=6.63×10−34×3.00×108632.8×10−9=1.989×10−256.328×10−7=3.14×10−19 JE = \frac{6.63\times10^{-34}\times3.00\times10^{8}}{632.8\times10^{-9}} = \frac{1.989\times10^{-25}}{6.328\times10^{-7}} = 3.14\times10^{-19}\text{ J}

Recompute as a check, going via the frequency first: f=cλ=3.00×1086.328×10−7=4.74×1014 Hzf=\dfrac{c}{\lambda}=\dfrac{3.00\times10^{8}}{6.328\times10^{-7}}=4.74\times10^{14}\text{ Hz}, so E=hf=6.63×10−34×4.74×1014=3.14×10−19 JE=hf=6.63\times10^{-34}\times4.74\times10^{14}=3.14\times10^{-19}\text{ J}. Both routes agree.

Step 2: Convert the energy to electron-volts

One electron-volt is the energy gained by an electron accelerated through a potential difference of 1 V, so 1 eV=1.60×10−19 J1\text{ eV}=1.60\times10^{-19}\text{ J}. Dividing the photon energy by ee:

EeV=Ee=3.14×10−191.60×10−19=1.96 eVE_{eV} = \frac{E}{e} = \frac{3.14\times10^{-19}}{1.60\times10^{-19}} = 1.96\text{ eV}

Why the other options are wrong

  • B (0.509 eV0.509\text{ eV}): comes from inverting the conversion, calculating e/Ee/E instead of E/eE/e.
  • C (3.14×10−19 eV3.14\times10^{-19}\text{ eV}): this is the correct energy in joules, simply relabelled with the wrong unit. The eV conversion was never applied.
  • D (196 eV196\text{ eV}): comes from a power-of-ten slip converting 632.8 nm632.8\text{ nm} to metres (e.g. using 6.328×10−9 m6.328\times10^{-9}\text{ m} shifted to 6.328×10−11 m6.328\times10^{-11}\text{ m}), which shrinks λ\lambda by a factor of 100 and so inflates EE by the same factor.

Final answer

  • The photon energy is 1.96 eV\boxed{1.96}\text{ eV}, option A.