Question

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

• A. 91 nm
• B. 192 nm
• C. 406 nm
• D. $9.1\times {10}^{-8}nm$

Answer 1

Answer: (A)

$\frac{1}{\lambda }={\bar{v}}_H={\bar{R}}_H\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
$=1.097\times {10}^7\left[\frac{1}{1^2}-\frac{1}{{\infty }^2}\right]$
$\therefore$ $\lambda =\frac{1}{1.097\times {10}^7}m$
$9.11\times {10}^{-8}m$
$91.1\times {10}^{-9}m$
$=91.1nm$ $(1nm={10}^{-9}m)$