COMEDK · Physics · 28. Atomic Physics
The wavelength of the second line of Balmer series is \(486.4 \mathrm{~nm}\). What is the wavelength of the first line of Lyman series?
- A \(78.8 \mathrm{~nm}\)
- B \(121.6 \mathrm{~nm}\)
- C \(418.2 \mathrm{~nm}\)
- D \(610.5 \mathrm{~nm}\)
Answer & Solution
Correct Answer
(B) \(121.6 \mathrm{~nm}\)
Step-by-step Solution
Detailed explanation
The Rydberg formula for the wavelength \(\lambda\) of a spectral line in a hydrogen-like atom is given by \(\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right) Z^2\). For hydrogen, \(Z = 1\). The second line of the Balmer series corresponds to the…
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