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Difficulty: MediumEnergy Levels and Atomic Spectra

An electron in a hydrogen atom makes a transition from the third energy level (n=3n = 3) to the second energy level (n=2n = 2). If the energy of the electron at n=3n = 3 is −1.51 eV-1.51\text{ eV} and at n=2n = 2 is −3.40 eV-3.40\text{ eV}, what is the frequency of the emitted photon? (h=6.6×10−34 J⋅sh = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}, 1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J})

  1. 4.58×1014 Hz4.58 \times 10^{14}\text{ Hz}Answer
  2. B
    2.86×1033 Hz2.86 \times 10^{33}\text{ Hz}
  3. C
    1.19×1015 Hz1.19 \times 10^{15}\text{ Hz}
  4. D
    8.24×1014 Hz8.24 \times 10^{14}\text{ Hz}

Answer

The frequency of the emitted photon is 4.58×1014 Hz4.58 \times 10^{14}\text{ Hz}.
When an electron transitions from a higher energy state to a lower energy state, it emits a photon whose energy equals the difference between the two levels: ΔE=E3−E2=1.89 eV\Delta E = E_3 - E_2 = 1.89\text{ eV}. Converting 1.89 eV1.89\text{ eV} to Joules gives 3.024×10−19 J3.024 \times 10^{-19}\text{ J}. Dividing by Planck's constant (6.6×10−34 J⋅s6.6 \times 10^{-34}\text{ J}\cdot\text{s}) gives a frequency of 4.58×1014 Hz4.58 \times 10^{14}\text{ Hz}.

Step-by-Step Solution

1
Calculate the energy difference between the initial and final energy levels.
ΔE=E3−E2=−1.51 eV−(−3.40 eV)=1.89 eV\Delta E = E_3 - E_2 = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}
The energy of the emitted photon equals the difference in energy between the two states.
2
Convert the energy change from electron-volts (eV) to Joules (J).
ΔE=1.89×1.6×10−19 J=3.024×10−19 J\Delta E = 1.89 \times 1.6 \times 10^{-19}\text{ J} = 3.024 \times 10^{-19}\text{ J}
SI units (Joules) are required to calculate frequency using Planck's constant in J⋅s\text{J}\cdot\text{s}.
3
Apply the photon energy formula E=hfE = hf to find the frequency ff.
f=ΔEh=3.024×10−19 J6.6×10−34 J⋅s=4.5818×1014 Hz≈4.58×1014 Hzf = \frac{\Delta E}{h} = \frac{3.024 \times 10^{-19}\text{ J}}{6.6 \times 10^{-34}\text{ J}\cdot\text{s}} = 4.5818 \times 10^{14}\text{ Hz} \approx 4.58 \times 10^{14}\text{ Hz}
Dividing energy by Planck's constant yields the photon frequency.

Key Concept

Photon Emission and Energy Level Transition
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Topics

Energy Levels and Atomic Spectra

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