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Quantum Physics: Question 8

Syllabus 22.4

Multiple choice A2 1 mark

An isolated atom of a certain gas has two relevant electron energy levels: E1=−8.20 eVE_1=-8.20\text{ eV} (the ground state) and E2=−3.00 eVE_2=-3.00\text{ eV}. A ground-state atom absorbs a photon, exciting an electron from E1E_1 to E2E_2.

What is the wavelength of the absorbed photon?

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

Step 1: Find the transition energy

When an electron is excited from E1E_1 to E2E_2, the atom absorbs a photon whose energy exactly equals the gap between the two levels: ΔE=E2−E1=−3.00−(−8.20)=5.20 eV\Delta E = E_2-E_1 = -3.00-(-8.20) = 5.20\text{ eV}

Converting to joules: ΔE=5.20×1.60×10−19=8.32×10−19 J\Delta E = 5.20\times1.60\times10^{-19} = 8.32\times10^{-19}\text{ J}

Step 2: Find the wavelength of the absorbed photon

Since ΔE=hf=hcλ\Delta E=hf=\dfrac{hc}{\lambda}, rearranging gives: λ=hcΔE\lambda = \frac{hc}{\Delta E}

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 ΔE=8.32×10−19 J\Delta E=8.32\times10^{-19}\text{ J}: λ=6.63×10−34×3.00×1088.32×10−19=1.989×10−258.32×10−19=2.39×10−7 m\lambda = \frac{6.63\times10^{-34}\times3.00\times10^{8}}{8.32\times10^{-19}} = \frac{1.989\times10^{-25}}{8.32\times10^{-19}} = 2.39\times10^{-7}\text{ m}

Recompute as a check, via the frequency first: f=ΔE/h=8.32×10−19/6.63×10−34=1.255×1015 Hzf=\Delta E/h=8.32\times10^{-19}/6.63\times10^{-34}=1.255\times10^{15}\text{ Hz}, so λ=c/f=3.00×108/1.255×1015=2.39×10−7 m\lambda=c/f=3.00\times10^{8}/1.255\times10^{15}=2.39\times10^{-7}\text{ m}. Both routes agree.

Why the other options are wrong

  • B (7.97×10−16 m7.97\times10^{-16}\text{ m}): comes from omitting the factor of cc, calculating λ=h/ΔE\lambda=h/\Delta E instead of λ=hc/ΔE\lambda=hc/\Delta E.
  • C (1.11×10−7 m1.11\times10^{-7}\text{ m}): comes from adding the magnitudes of the two energy levels instead of subtracting, using ΔE=∣−8.20∣+∣−3.00∣=11.20 eV\Delta E=|{-8.20}|+|{-3.00}|=11.20\text{ eV}.
  • D (3.83×10−26 m3.83\times10^{-26}\text{ m}): comes from forgetting to convert the transition energy from eV to joules, using ΔE=5.20\Delta E=5.20 directly as if it were already in joules.

Final answer

  • The absorbed photon has wavelength 2.39×10−7 m\boxed{2.39\times10^{-7}}\text{ m} (i.e. 239 nm239\text{ nm}), option A.