Calculate the radius of the first orbit of hydrogen atom. Show that the velocity of electron in the first orbit is $\frac{1}{137} times the velocity of light .$

Solution

Radius of any orbit is given by the formula,

$r = \frac{n^{2} h^{2}}{4 π^{2} {mKZe}^{2}}$

We have, $n = 1 for 1 st orbit$ and,
$h = 6.6 \times 10^{- 34} Js m = 9 \times 10^{- 31} kg K = 9 \times 10^{9} {Nm}^{2} C^{- 2} Z = 1 for hydrogen, e = 1.6 \times 10^{- 19} coulomb$

Therefore, putting the values in the formula,

we get,
$r = 0.53 \times 10^{- 10} m$ , is the radius of the first orbit of hydrogen atom.

Now,
Velocity of electron in the first orbit is,
$v = \frac{2 {πKe}^{2}}{nh} = \frac{c}{n} (\frac{2 π {Ke}^{2}}{ch})$

$= \frac{c}{1} \times 2 \times \frac{22}{7} \times \frac{9 \times 10^{9} \times (1.6 \times 10^{- 19})^{2}}{3 \times 10^{8} \times 6.6 \times 10^{- 34}}$

ie., $v = \frac{1}{137} c .$

Hence proved.