Question

What is the maximum number of electrons present in an orbit whose angular momentum is $\frac{3h}{\pi }$?

• A. $72$
• B. $64$
• C. $36$
• D. $81$

Answer 1

The correct option is A

$72$

We know that, Angular momentum = $\frac{nh}{2\pi }$
$\therefore \frac{nh}{2\pi }=\frac{3h}{\pi }$
i.e $n=6$
Hence maximum number of electrons present = $2n^2$
= $2(6\times 6)$ = $72$.