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Quantum Numbers

The angular m...

Question

The angular momentum of an electron in a Bohr orbit of $H$ -atom is $4.2178 \times 10^{- 34} k g m^{2} / s e c$ . Calculate wavelength of the spectral line emitted when an electron falls from this level to the next lower level.

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Solution

We know, $m v r = n \frac{h}{2 π}$

$4.2178 \times 10^{- 34} = n \times \frac{6.626 \times 10^{- 34}}{2 \times 3.14}$

$∴$ $n = 4$

$\frac{1}{λ} = R_{H} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

$= 109678 [\frac{1}{3^{2}} - \frac{1}{4^{2}}]$
$λ = 1.8 \times 10^{- 4} c m$ .

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