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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, $e$ is the charge on the electron and $\varepsilon _{0}$ is the vacuum permittivity, the speed of the electron is
Option: 1
$0$

Option: 2
$\frac{e}{\sqrt{\varepsilon_0 a_0 m}}$

Option: 3
$\frac{e}{\sqrt{4 \pi \varepsilon_0 a_0 m}}$

Option: 4
$\frac{\sqrt{4 \pi \varepsilon_0 a_0 m}}{e}$

Answers (1)

best_answer

$\frac{m v^2}{a_0}=\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{a_0^2} \Rightarrow v=\frac{e}{\sqrt{4 \pi \varepsilon_0 a_0 m}}$

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