According to Bohr's theory, the moment of momentum of an electron revolving in $4^{\text {th }}$ orbit of…
According to Bohr's theory, the moment of momentum of an electron revolving in $4^{\text {th }}$ orbit of hydrogen atom is:
- $\frac{h}{\pi}$
- $\frac{h}{2 \pi}$
- $8 \frac{h}{\pi}$
- $2 \frac{h}{\pi}$
Solution
Moment of momentum is $\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{P}}$
$\begin{aligned}
& \overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} \times \mathrm{m} \overrightarrow{\mathrm{v}} \\
& \mathrm{L}=\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}=\frac{4 \mathrm{~h}}{2 \pi}=\frac{2 \mathrm{~h}}{\pi}
\end{aligned}$
Asked in: JEE Main 2024 (04 Apr Shift 2)
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