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  • Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength . The value of principal quantum number <img alt="\m

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength \lambda. The value of principal quantum number \mathrm{' n '} of the excited state will be : (R : Rydberg constant)
 

Option: 1

\sqrt{\frac{\lambda \mathrm{R}}{\lambda-1}}


Option: 2

\sqrt{\frac{\lambda \mathrm{R}}{\lambda \mathrm{R}-1}}


Option: 3

\sqrt{\frac{\lambda}{\lambda \mathrm{R}-1}}


Option: 4

\sqrt{\frac{\lambda R^{2}}{\lambda R-1}}


Answers (1)

best_answer

\mathrm{\frac{1}{\lambda}=R z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]}
For hydrogen atom, \mathrm{z=1}

\mathrm{\frac{1}{\lambda}=R\left[\frac{1}{1}-\frac{1}{n^2}\right] }

\mathrm{\frac{1}{\lambda R}=1-\frac{1}{n^2} }

\mathrm{\frac{1}{n^2}=1-\frac{1}{\lambda R}=\frac{(\lambda R-1)}{\lambda R} }

\mathrm{n=\sqrt{\frac{\lambda R}{(\lambda R-1)}}}

Hence (2) is correct option

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