Question

The orbital angular momentum of an electron is $\sqrt{3}\frac{h}{\pi }$, Which of the following of angular momentum of this electron revolving in unknown Bohr orbit?

• A. $\frac{h}{2\pi }$
• B. $\frac{h}{\pi }$
• C. $3\frac{h}{\pi }$
• D. none of the above

Answer 1

The correct option is C $3\frac{h}{\pi }$

Solution :

Orbital angular momentum $=\sqrt{l(1+l)}\frac{h}{2\pi }$ (l = azimulthal quantum no)

Given $O.A.M=\sqrt{3}\frac{h}{\pi }$

$\therefore \sqrt{L(1+1)}\times \frac{h}{2\pi }=\sqrt{3}\frac{h}{\pi }$

$\Rightarrow \frac{L(1+1)}{4}=3\Rightarrow l(1+1)=12$

We have : $l^2+l-12=0\Rightarrow (l+2)(l-3)=0$

Possible values of $L=-4,3.(-4$ not allowed as $L<0)$

$\therefore L=3$

For $L=3$, Possible values of $n=4,5,6$

Thus, possible angular momentum $=\frac{4h}{2\pi },\frac{5h}{2\pi },\frac{6h}{2\pi },\frac{7h}{2\pi },...$

From the all available Options, $\frac{3h}{\pi }$ is correct answer which is equivalent to $\frac{bh}{2\pi }$ (at $n=6$ )

Correct Ans: (C)