In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the centre at the proton. The proton itself is assumed to be fixed in an inertial frame. The centripetal force is provided by the Coulomb attraction. In the ground state, the electron goes round the proton in a circle of radius 5.3 × 10-11 m. Find the speed of the electron in the ground state. Mass of the electron = 9.1 × 10-31 kg and charge of the electron = 1.6 × 10-19 C.

It is given that the centripetal force is being provided by coulomb attractions.

⇒ (where all the symbols have their usual meanings)


⇒ v2 = kq2/mr


It is given that,


Mass of electron, m = 9.1 × 10-31 kg


Radius of the circular path, r = 5.3 × 10-11 m


k=1/(4πε0) = 9×109


q1 = charge on an electron= q2 = charge on the proton = q = 1.6 × 10-19 C


⇒ v2 = 9×109×1.6×10-19×1.6×10-19/(9.31×10-31×5.3×10-11)


⇒ v2 = 4.67 × 1012


⇒ v = 2.16 × 106 m/s


⇒ v ≈ 2.2 × 106 m/s


∴ Speed of the electron in ground state = 2.2 × 106 m/s


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