Question

Ratio of angular momentum for an electron in $3^{rd}$ orbit of $L{i}^{++}$ and $5^{th}$ orbit of $H{e}^{+}$ is

• A. $\frac{3}{5}$
• B. $\frac{9}{25}$
• C. $1$
• D. $\frac{5}{3}$

Answer 1

The correct option is A $\frac{3}{5}$

Angular momentum $L$ of an electron in hydrogen atom is given by $\frac{nh}{2\pi }$ .

And as $L{i}^{++}$ and $H{e}^{+}$ are similar to hydrogen atom, we can use the same formula.
So,
$\frac{L_1}{L_2}=\frac{\frac{3h}{2\pi }}{\frac{5h}{2\pi }}=\frac{3}{5}$