Question

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

A
L=hπ
B
L=4hπ
C
L=3hπ
D
L=5hπ

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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