The wavelength of radiation emitted when an electron jump from $n = 4$ to $n = 2$ energy levels of hydrogen atom is (Given: Rydberg constant= $1.097 \times 10^{7} m^{- 1}$ )

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$486 nm$

$252 nm$

$890 nm$

$342 nm$

answer is C.

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Detailed Solution

$\frac{1}{λ} = R_{\infty} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = (1.097 \times 10^{7} m^{- 1}) (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = (1.097 \times 10^{7} {mol}^{- 1}) (\frac{3}{16})$

Thus the value this

$λ = 4.86 \times 10^{- 7} m = 486 nm$

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