Question

Show that the speed of an electron in 1st Bohr orbit is $\frac{c}{137}$, where c is the speed of light.

Answer 1

Let v be the speed of the electron (mass m, charge e) in the nth orbit whose radius is r. The charge on hydrogen nucleus is +e. The centripetal force is provided by the electrostatic attraction.
$\therefore \frac{m{v}^2}{r}=\frac{1}{4\pi {ϵ}_0}\frac{e^2}{r^2}\dots \dots \left(i\right)$
According to Bohr's first postulate, $mvr=\frac{nh}{2\pi }\dots \dots \left(ii\right)$
Dividing Eq. (i) by Eq. (ii), $v=\frac{e^2}{2nh{ϵ}_0}$
The velocity in the first orbit (n = 1) is given by
$v_1=\frac{(1.6\times {10}^{-19}{)}^2}{2\times 1\times (6.6\times {10}^{-34})(8.854\times {10}^{-12})}=2.19\times {10}^{6}m{s}^{-1}$
Now, $\frac{v_1}{c}=\frac{2.19\times {10}^6}{3\times {10}^8}=\frac{1}{137}\therefore v_1=\frac{c}{137}$