Question

The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state $ 1 $ , 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)

$\frac{1}{\lambda}=\bar{v}_{H}=\bar{R}_{H}\left[\frac{1}{n^{2}_{1}}-\frac{1}{n^{2}_{2}}\right]$
$=1.097\times10^{7}\left[\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right]$
$\therefore \lambda=\frac{1}{1.097\times10^{7}}\,m$
$=9.11\times10^{-8}\,m$
$=91.1\times10^{-9}\,m$
$=91.1\,nm\quad\left(1\,nm=10^{-9}\,m\right)$