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Question

What is the angular momentum of an electron in Bohr's hydrogen atom whose energy is $$-3.4\space eV$$?


A
L=hπ
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B
L=4hπ
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C
L=3hπ
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D
L=5hπ
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Solution

The correct option is A $$L = \displaystyle\frac{h}{\pi}$$
First, we need to identify the quantum number of the energy level.
As $$E = - \displaystyle\frac{13.6}{n^2}$$, we have
$$\quad -3.4 = -\displaystyle\frac{13.6}{n^2}$$ or $$n = 2$$
The angular momentum quantization gives, $$L = mvr = \displaystyle\frac{nh}{2\pi}$$
Substituting $$n=2$$, we get $$L = \displaystyle\frac{2h}{2\pi} = \displaystyle\frac{h}{\pi}$$

Physics

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