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

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$9.1 \times 10^{- 8} n m$

192 nm

406 nm

91 nm

answer is C.

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Detailed Solution

$\frac{1}{λ} = R (\frac{1}{n_{1}^{1}} - \frac{1}{n_{2}^{2}})$

$\frac{1}{λ} = 1.097 \times 10^{7} (\frac{1}{1} - \frac{1}{\infty}) = 1.097 \times 10^{7}$

$λ = 91.15 \times 10^{- 9} m \approx 91 n m$

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