The angular momentum for the electron in Bohr's orbit is L. If the...

The angular momentum for the electron in Bohr's orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be:

Options

A $\frac{\mathrm{L}}{2}$
B
zero
C
L
D
2L

Correct answer

C.
L

Step-by-step solution

From the Bohr's quantization rule
Angular momentum

\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}, \mathrm{L}_1=\frac{1. \mathrm{h}}{2 \pi}=\mathrm{L}

(n=1

for first orbit)<br />in (n=2

for second orbit)

L_2=\frac{2 h}{2 \pi}=2 L

Hence, change in angular momentum

\mathrm{L}_2-\mathrm{L}_1=2 \mathrm{ L}-\mathrm{L}=\mathrm{L}

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