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 :
Select the correct option:
A
$\frac{\mathrm{L}}{2}$
B
zero
C
L
D
2 L
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Sol.
$$
\begin{aligned}
& \mathrm{L}=m v r \quad \mathrm{r} \propto \mathrm{n}^2, \quad \mathrm{v} \propto \frac{1}{\mathrm{n}} \\
& \therefore \mathrm{~L} \propto \mathrm{n}
\end{aligned}
$$
Also, $\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}$, Bohr orbit is, $\mathrm{L}_1=\mathrm{L}=\frac{1 \cdot h}{2 \pi}$
$$
\begin{aligned}
& L_2=2[L]=2 L \\
& L_2=\frac{2 h}{2 \pi}
\end{aligned}
$$
So, change $=\mathrm{L}_2-\mathrm{L}_1=2 \mathrm{~L}-\mathrm{L}=\mathrm{L}$
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