MCAT Physics Question 124: Answer and Explanation

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Question: 124

4. When a hydrogen atom electron falls to the ground state from the n = 2 state, 10.2 eV of energy is emitted. What is the wavelength of this radiation? (Note: 1 eV = 1.60 × 10-19 J, and h = 6.626 × 10-34 J·s.)

  • A. 5.76 × 10-9 m
  • B. 1.22 × 10-7 m
  • C. 3.45 × 10-7 m
  • D. 2.5 × 1015 m

Correct Answer: B

Explanation:

To solve this question correctly, one must be careful with the units. First, convert 10.2 eV to joules:

Next, to determine the wavelength of the radiation, first find the frequency:

Lastly, from the wave equation c = f λ, we can calculate the wavelength of the radiation:

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