Question

The number of revolution per second made by an electron in the first Bohr orbit of the hydrogen atom is of the order of,

• A. ${10}^{20}$
• B. ${10}^{19}$
• C. ${10}^{17}$
• D. ${10}^{15}$

Answer 1

The correct option is D ${10}^{15}$

Angular Momentum of an electron is given as, $mvr=\frac{nh}{2\pi }$
for first orbit, $n=1$
$mvr=\frac{h}{2\pi }$
$\Rightarrow m{r}^2\omega =\frac{h}{2\pi }$
$\Rightarrow m\left(2\pi f\right)r^2=\frac{h}{2\pi }[\because \omega =2\pi f]$
$\Rightarrow f=\frac{h}{4{\pi }^{2}m{r}^2}$
$=\frac{6.63\times {10}^{-34}}{4\times (3.14{)}^2\times 9.1\times {10}^{-31}\times (0.53\times {10}^{-10}{)}^2}$

$\Rightarrow f\approx 6.6\times {10}^{15}\text{rev/sec}$

Hence, $\left(D\right)$ is the correct answer.