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Question

Speed of electron in $\mathrm{n}^{	ext {th }}$ orbit of hydrogen atom

$\mathrm{v}=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{nh}}$

In grou

Vedclass · Accepted Answer

$1 / 137$

Vedclass · Answer

Speed of electron in $\mathrm{n}^{	ext {th }}$ orbit of hydrogen atom

$\mathrm{v}=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{nh}}$

In ground state $\mathrm{n}=1 \Rightarrow \frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{h}}$

$\Rightarrow \frac{\mathrm{v}}{\mathrm{c}}=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{ch}}=\frac{\left(1.6 	imes 10^{-19}ight)^{2}}{2 	imes 8.85 	imes 10^{-12} 	imes 3 	imes 10^{8} 	imes 6.6 	imes 10^{-34}}$

$=\frac{1}{137}$

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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