Question

The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state, would be: [Rydberg constant $=1.097\times {10}^7{\text{m}}^{-1}]$

• A. $406\text{nm}$
• B. $192\text{nm}$
• C. $9.1\times {10}^{-4}\text{nm}$
• D. $91\text{nm}$

Answer 1

The correct option is D $91\text{nm}$

$\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\frac{1}{\lambda }=R_H[\frac{1}{1^2}-\frac{1}{{\infty }^2}]$

$\therefore \lambda =\frac{1}{R_H}$

$\Rightarrow \lambda =\frac{1}{1.097\times {10}^7}\text{m}$

$\Rightarrow \lambda =9.1\times {10}^{-8}m$

$\Rightarrow \lambda =91\times {10}^{-9}\text{m}$

$\Rightarrow \lambda =91\text{nm}$
Hence, option A is correct.