First slide

Bohr Model of the Hydrogen atom.

Question

If a hydrogen atom emits a photon of energy 12.75 eV and returns to its ground state, change in its orbital angular momentum is

Moderate

A

$\frac{3 h}{2 π}$
B

$\frac{2 h}{π}$
C

$\frac{h}{2 π}$
D

$\frac{h}{π}$

Solution

(1)
$\begin{array}{l} Δ E = 13.6 (1 - \frac{1}{n^{2}}) \\ ∴ 12.75 = 13.6 (1 - \frac{1}{n^{2}}) \Rightarrow n = 4 \\ ∴ Δ L = 4. \frac{h}{π} - 1. \frac{h}{2 π} = \frac{3 h}{2 π} \end{array}$

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