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The ratio of speed of an electron in the ground state in the Bohr's first orbit of hydrogen atom to velocity of light (c) is (h $=$ Planck's constant, $\epsilon_{0}=$ permittivity of free space, $\mathrm{e}=$ charge on electron $)$
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Verified Answer
The correct answer is:
$\frac{e^{2}}{2 \in_{0} h c}$
The velocity of electron in Bohr's first orbit is
$$
\begin{aligned}
\mathrm{V} &=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{~h}} \\
\therefore \quad \frac{\mathrm{V}}{\mathrm{c}} &=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{hc}}
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{V} &=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{~h}} \\
\therefore \quad \frac{\mathrm{V}}{\mathrm{c}} &=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{hc}}
\end{aligned}
$$
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