JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2022
- Hard
- MCQ
Hydrogen atom from excited state comes to the ground by emitting a photon of wavelength \(\lambda\). The value of principal quantum number ' \(n\) ' of the excited state will be: (\(R\): Rydberg constant)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{\lambda R}{\lambda R-1}}\)
Step-by-step Solution
Detailed explanation
\(\frac{- Rch }{( n )^{2}}+\frac{ Rch }{1}=\frac{ hc }{\lambda}\)
\(\frac{- R }{ n ^{2}}+ R =\frac{1}{\lambda}\)
\(R -\frac{1}{\lambda}=\frac{ R }{ n ^{2}}\)
\(\frac{\lambda R -1}{\lambda}=\frac{ R }{ n ^{2}}\)
\(n ^{2}=\frac{\lambda R }{\lambda R -1} \Rightarrow n =\sqrt{\frac{\lambda R }{\lambda R -1}}\)
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