Hydrogen atom from exited state comes to the ground state by emitting a photon of wavelength $\lambda .$ The value of principal quantum number ‘n’ of the excited state will be

Correct Option - 2


Detailed Solution:

According to definition of wave number, we can write

$\frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\Rightarrow 1-\frac{1}{n^2}=\frac{1}{\lambda R}$$$

$\Rightarrow \frac{1}{n^2}=1-\frac{1}{\lambda R}=\frac{\lambda R-1}{\lambda R}\Rightarrow n=\sqrt[]{\frac{\lambda R}{\lambda R-1}}$

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