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What is the energy of a photon emitted when an electron transitions from n=3 to

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Question: What is the energy of a photon emitted when an electron transitions from n=3 to n=2 in a hydrogen atom? (Rydberg constant R = 1.097 x 10^7 m^-1) (2021)

Options:

  1. 1.89 eV
  2. 3.40 eV
  3. 4.86 eV
  4. 5.13 eV

Correct Answer: 3.40 eV

Exam Year: 2021

Solution: 

Energy = 13.6 eV * (1/n1^2 - 1/n2^2) = 13.6 * (1/2^2 - 1/3^2) = 3.40 eV.

What is the energy of a photon emitted when an electron transitions from n=3 to

Practice Questions

Q1
What is the energy of a photon emitted when an electron transitions from n=3 to n=2 in a hydrogen atom? (Rydberg constant R = 1.097 x 10^7 m^-1) (2021)
  1. 1.89 eV
  2. 3.40 eV
  3. 4.86 eV
  4. 5.13 eV

Questions & Step-by-Step Solutions

What is the energy of a photon emitted when an electron transitions from n=3 to n=2 in a hydrogen atom? (Rydberg constant R = 1.097 x 10^7 m^-1) (2021)
  • Step 1: Identify the initial and final energy levels of the electron. Here, n1 = 2 (final level) and n2 = 3 (initial level).
  • Step 2: Use the formula for the energy of a photon emitted during an electron transition: Energy = 13.6 eV * (1/n1^2 - 1/n2^2).
  • Step 3: Substitute the values of n1 and n2 into the formula: Energy = 13.6 eV * (1/2^2 - 1/3^2).
  • Step 4: Calculate 1/2^2, which is 1/4 = 0.25.
  • Step 5: Calculate 1/3^2, which is 1/9 ≈ 0.1111.
  • Step 6: Subtract the two results: 0.25 - 0.1111 = 0.1389.
  • Step 7: Multiply the result by 13.6 eV: Energy = 13.6 eV * 0.1389 ≈ 1.89 eV.
  • Step 8: Round the final answer to two decimal places if necessary.
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