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(C) The speed of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $v = \frac{e^2}{2\varepsilon_0 nh}$.For the ground state

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$1/137$

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(C) The speed of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $v = \frac{e^2}{2\varepsilon_0 nh}$.For the ground state,the principal quantum number $n = 1$.Substituting $n = 1$,we get the speed $v = \frac{e^2}{2\varepsilon_0 h}$.To find the ratio of the speed of the electron $(v)$ to the speed of light $(c)$,we calculate $\frac{v}{c} = \frac{e^2}{2\varepsilon_0 ch}$.Substituting the physical constants: $e = 1.6 	imes 10^{-19} \ C$,$\varepsilon_0 = 8.85 	imes 10^{-12} \ F/m$,$c = 3 	imes 10^8 \ m/s$,and $h = 6.63 	imes 10^{-34} \ J\cdot s$.$\frac{v}{c} = \frac{(1.6 	imes 10^{-19})^2}{2 	imes 8.85 	imes 10^{-12} 	imes 3 	imes 10^8 	imes 6.63 	imes 10^{-34}} \approx \frac{1}{137}$.Thus,the ratio is $1/137$.

The ratio of the speed of the electrons in the ground state of hydrogen to the speed of light in vacuum is

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