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An electron in a hydrogen atom jumps from the third excited state ($n = 4$) to the ground state ($n = 1$). What is the wavelength $\lambda$ of the emitted photon? (Take $R = 1.097 \times 10^7$ m$^{-1}$)
- A.$97.2$ nm
- B.$102.6$ nm
- C.$121.6$ nm
- D.$91.2$ nm
Show correct answer & step-by-step solution
Correct answer: A — $97.2$ nm
Solution
Using the Rydberg formula , with and , we get . Solving for gives approximately nm.
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