The ratio of speed of an electron in ground state in Bohrs first orbit of hydrogen atom to velocity of light in air is

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$\frac{2 ε_{0} hc}{e^{2}}$

$\frac{2 e^{2} ε_{0}}{hc}$

$\frac{e^{3}}{2 ε_{0} hc}$

$\frac{e^{2}}{2 ε_{0} hc}$

answer is A.

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Detailed Solution

Speed of electron in nth orbit (in CGS) $v_{n} = \frac{2 {πZe}^{2}}{nh}$ (k = 1)
For first orbit H₂; n = 1 and Z = 1
So $v = \frac{2 {πe}^{2}}{h}$ $\Rightarrow$ $\frac{v}{c} = \frac{2 {πe}^{2}}{hc}$

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