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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)

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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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