Question

The maximum angular speed of the electron of a hydrogen atom in a stationary orbit is

• A. $6.2\times {10}^{15}rad{s}^{-1}$
• B. $4.1\times {10}^{16}rad{s}^{-1}$
• C. $24\times {10}^{10}rad{s}^{-1}$
• D. $9.2\times {10}^{6}rad{s}^{-1}$

Answer 1

The correct option is B $4.1\times {10}^{16}rad{s}^{-1}$

Angular velocity is $\omega =\frac{v}{r}$
$\omega =\frac{2.16\times {10}^6}{0.529\times {10}^{-10}}\frac{Z^2}{n^3}$

which is maximum for $n=1$, and given $Z=1$

$\therefore \omega =\frac{2.16\times {10}^6}{0.529\times {10}^{-10}}=4.08\times {10}^{16}rad/s$