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? (2019)

Options:

Correct Answer: 1.89 eV

Exam Year: 2019

Solution:

The energy difference can be calculated using the formula E = 13.6 eV (1/n1² - 1/n2²). For n1=2 and n2=3, E = 13.6 eV (1/2² - 1/3²) = 1.89 eV.

What is the energy of a photon emitted when an electron transitions from n=3 to n=2 in a hydrogen atom? (2019)

What is the energy of a photon emitted when an electron transitions from n=3 to n=2 in a hydrogen atom? (2019)

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 energy difference: E = 13.6 eV (1/n1² - 1/n2²).
Step 3: Substitute the values into the formula: E = 13.6 eV (1/2² - 1/3²).
Step 4: Calculate 1/2², which is 1/4 = 0.25.
Step 5: Calculate 1/3², which is 1/9 ≈ 0.1111.
Step 6: Find the difference: 0.25 - 0.1111 = 0.1389.
Step 7: Multiply the difference by 13.6 eV: E = 13.6 eV * 0.1389 ≈ 1.89 eV.
Step 8: Conclude that the energy of the photon emitted is approximately 1.89 eV.

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