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In the Bohr model of a hydrogen atom, the centripetal force is furnished by the coulomb attraction between the proton and the electron. If $a_{0}$ is the radius of the ground state orbit, $m$ is the mass and e is charge on the electron and $\varepsilon_{0}$ is the vacuum permittivity, the speed of the electron is
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The correct answer is:
$\frac{\mathrm{e}}{\sqrt{4 \pi \varepsilon_{0} \mathrm{a}_{0} \mathrm{~m}}}$
Centripetal force $=$ force of attraction of nucleus on electron.
$$
\frac{m v^{2}}{a_{0}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{e^{2}}{a_{0}^{2}} \Rightarrow v=\frac{e}{\sqrt{4 \pi \varepsilon_{0} m a_{0}}}
$$
$$
\frac{m v^{2}}{a_{0}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{e^{2}}{a_{0}^{2}} \Rightarrow v=\frac{e}{\sqrt{4 \pi \varepsilon_{0} m a_{0}}}
$$
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