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JEE Mains · Physics · STD 12 - 12. atoms

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)

  1. A \(\sqrt{\frac{\lambda R}{\lambda-1}}\)
  2. B \(\sqrt{\frac{\lambda R}{\lambda R-1}}\)
  3. C \(\sqrt{\frac{\lambda}{\lambda R-1}}\)
  4. D \(\sqrt{\frac{\lambda R^{2}}{\lambda R-1}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\sqrt{\frac{\lambda R}{\lambda R-1}}\)

Step-by-step Solution

Detailed explanation

\(\frac{- Rch }{( n )^{2}}+\frac{ Rch }{1}=\frac{ hc }{\lambda}\) \(\frac{- R }{ n ^{2}}+ R =\frac{1}{\lambda}\) \(R -\frac{1}{\lambda}=\frac{ R }{ n ^{2}}\) \(\frac{\lambda R -1}{\lambda}=\frac{ R }{ n ^{2}}\)…
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