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 $
$ =91.1\,nm $ $ (1\,nm={{10}^{-9}}\,m) $