Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. According to Bohr's theory angular momentum of an electron in $6^{\text {th }}$ orbit is

Structure of Atom

Solution:

$mvr =\frac{ nh }{2 \pi}$
( $n=$ number of shell)
Angular momentum
for $6^{\text {th }} \text { shell }=\frac{6 h}{2 \pi}=\frac{3 h}{\pi}$