MCQ
The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state $1$, would be ............... $\mathrm{nm}$ (Rydberg constant $ = \,1.097\, \times \,{10^7}\,{m^{ - 1}}$)
  • A
    $406$
  • B
    $192$
  • $91$
  • D
    $9.1\, \times \,{10^{ - 8}}$

Answer

Correct option: C.
$91$
c
(c) $\frac{1}{\lambda } = R\,\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]$

$\frac{1}{\lambda } = 1.097 \times {10^7}{m^{ - 1}}\left[ {\frac{1}{{{1^2}}} - \frac{1}{{{\infty ^2}}}} \right]$

$\therefore $ $\lambda = 91 \times {10^{ - 9}}\,m$

We know ${10^{ - 9}} = 1\,nm$ So $\lambda = 91\,nm$

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