Question

How many revolutions does an electron complete in one second in the first orbit of hydrogen atom?

• A. 6.5 x 10${}^{15}$
• B. 6.5 x 10${}^{13}$
• C. 1000
• D. 6.5 x 10${}^{14}$

Answer 1

Answer: (A)

6.5 x 10${}^{15}$

The orbital frequency is given by:
$\nu =\frac{v}{2\pi r}$ .....................(1)

The velocity of revolving electron is:
$v=\frac{e^2}{2nh{ϵ}_0}$

The radius of an orbit of revolution of electron is:
$r=\frac{n^2{h}^2{ϵ}_0}{\pi m{e}^2}$

Using these values in (1) we get:
$\nu =\frac{\frac{e^2}{2nh{ϵ}_0}}{2\pi \left(\frac{n^2{h}^2{ϵ}_0}{\pi m{e}^2}\right)}$

$\nu =\frac{1}{2\pi }\left(\frac{e^2}{2nh{ϵ}_0}\right)\left(\frac{\pi m{e}^2}{n^2{h}^2{ϵ}_0}\right)$

$\nu =\frac{m{e}^4}{4{ϵ}_0^2{n}^3{h}^3}$

For $n=1$,
$\nu =\frac{9.1\times {10}^{-31}\times (1.6\times {10}^{-19}{)}^4}{4\times (8.85\times {10}^{-12}{)}^2\times 1^3\times (6.6\times {10}^{-34}{)}^3}$

$\nu =\frac{59.63\times {10}^{-107}}{91179.3\times {10}^{-126}}$

$\nu =6.539\times {10}^{15}Hz$