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Difficulty: Very hardEnergy Levels and Atomic Spectra

A hydrogen atom initially in its ground state with energy E1=−13.6 eVE_1 = -13.6\text{ eV} absorbs a photon with energy 12.75 eV12.75\text{ eV}, raising the electron to an excited quantum level nn. The atom then undergoes de-excitation to lower energy levels. Taking Planck's constant h=6.6×10−34 J⋅sh = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}, speed of light c=3.0×108 m/sc = 3.0 \times 10^8\text{ m/s}, and 1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J}, what is the shortest wavelength of light emitted during any of the possible downward transitions?

  1. 9.7×10−8 m9.7 \times 10^{-8}\text{ m}Answer
  2. B
    1.55×10−26 m1.55 \times 10^{-26}\text{ m}
  3. C
    1.88×10−6 m1.88 \times 10^{-6}\text{ m}
  4. D
    1.02×10−7 m1.02 \times 10^{-7}\text{ m}

Answer

9.7×10−8 m9.7 \times 10^{-8}\text{ m}
The correct answer of 9.7×10−8 m9.7 \times 10^{-8}\text{ m} is obtained by finding that the absorbed photon of 12.75 eV12.75\text{ eV} promotes the electron to the n=4n = 4 state (E4=−0.85 eVE_4 = -0.85\text{ eV}). The shortest wavelength photon is emitted in the transition with the greatest energy difference, which is n=4→n=1n = 4 \rightarrow n = 1 (ΔE=12.75 eV\Delta E = 12.75\text{ eV}). Converting 12.75 eV12.75\text{ eV} to 2.04×10−18 J2.04 \times 10^{-18}\text{ J} and substituting into λ=hcΔE\lambda = \frac{hc}{\Delta E} yields 9.7×10−8 m9.7 \times 10^{-8}\text{ m}.

Step-by-Step Solution

1
Determine the energy of the excited state EnE_n
En=E1+ΔEabsorbed=−13.6 eV+12.75 eV=−0.85 eVE_n = E_1 + \Delta E_{\text{absorbed}} = -13.6\text{ eV} + 12.75\text{ eV} = -0.85\text{ eV}
Absorbing energy elevates the atom from its ground state energy to a higher energy level.
2
Identify the principal quantum number nn of the excited state
n=−13.6 eV−0.85 eV=16=4n = \sqrt{\frac{-13.6\text{ eV}}{-0.85\text{ eV}}} = \sqrt{16} = 4
For hydrogen, En=E1n2E_n = \frac{E_1}{n^2}.
3
Identify the transition giving the shortest wavelength
Transition from n=4→n=1n = 4 \rightarrow n = 1 with maximum energy change ΔEmax=12.75 eV\Delta E_{\text{max}} = 12.75\text{ eV}
Since λ=hcΔE\lambda = \frac{hc}{\Delta E}, the shortest wavelength occurs at maximum energy emission.
4
Convert energy to Joules and calculate wavelength
ΔE=12.75×1.6×10−19 J=2.04×10−18 J\Delta E = 12.75 \times 1.6 \times 10^{-19}\text{ J} = 2.04 \times 10^{-18}\text{ J}; λ=6.6×10−34×3.0×1082.04×10−18≈9.7×10−8 m\lambda = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{2.04 \times 10^{-18}} \approx 9.7 \times 10^{-8}\text{ m}
Applying the photon energy-wavelength relation E=hcλE = \frac{hc}{\lambda} in SI units.

Key Concept

Energy Levels and Atomic Spectra
Estimated Time:3m 0s
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Energy Levels and Atomic Spectra

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