Question

In the Bohr Model of a hydrogen atom, the centripetal force is furnished by the Coulomb attraction between the proton and the electron. If $a_o$ is the radius of the ground state orbit, $m$ is the mass, and $e$ is the charge on the electron, and ${ϵ}_o$ is the permittivity of vacuum, then the speed of the electron is:

• A. $0$
• B. $\frac{e}{\sqrt{{ϵ}_o{a}_{o}m}}$
• C. $\frac{e}{\sqrt{4\pi {ϵ}_o{a}_{o}m}}$
• D. $\sqrt{\frac{4\pi {ϵ}_o{a}_{o}m}{e}}$

Answer 1

The correct option is C $\frac{e}{\sqrt{4\pi {ϵ}_o{a}_{o}m}}$

The centripetal force is balanced by the coulombic force, hence,

$\frac{m{v}^2}{a_0}=\frac{1}{4\pi {ϵ}_0}\frac{e^2}{a_0^2}$(since $a_0$ is the radius of the first orbit where the speed is to be calculate)

$⟹v^2=\frac{e^2}{4\pi {ϵ}_0{a}_{0}m}$

$⟹v=\frac{e}{\sqrt{4\pi {ϵ}_0{a}_{0}m}}$