Question

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength λ. If r is the rydberg constant the principal quantum number n of the excited state is

Answer 1

Answer: The principal quantum number n of the excited state is :$n=\sqrt{\frac{\lambda R}{1+\lambda R}}$.

Explanation:

The initial state of the electron $n_i=n$

The final state of the electron after coming to the $n_f=1$

According to the Rydberg formula:

R =Rydberg constant= $1.097\times 10^{7} m^{-1}$

$\frac{1}{\lambda }=R[\frac{1}{n_{i}^2}-\frac{1}{n_{f}^2}]$

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

$n=\sqrt{\frac{\lambda R}{1+\lambda R}}$

The principal quantum number n of the excited state is :$n=\sqrt{\frac{\lambda R}{1+\lambda R}}$.

$\lambda$ = Photon's of wavelength