Q90 / 100 · BITSAT 2023 (Memory Based Paper 2) · Physics
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:
- A$\frac{\mathrm{L}}{2}$
- B
zero - Correct answerC
L - D
2L
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)
in ( $n=2$ for second orbit) $L_2=\frac{2 h}{2 \pi}=2 L$
Hence, change in angular momentum $\mathrm{L}_2-\mathrm{L}_1=2 \mathrm{~L}-$ $\mathrm{L}=\mathrm{L}$
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)
in ( $n=2$ for second orbit) $L_2=\frac{2 h}{2 \pi}=2 L$
Hence, change in angular momentum $\mathrm{L}_2-\mathrm{L}_1=2 \mathrm{~L}-$ $\mathrm{L}=\mathrm{L}$
Hence, the correct answer is option: 'C'.