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)

91 nm

Answer 2

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