Question

In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the center 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 around the proton in a circle of radius 5.3x10-11 m. Find the speed of the electron in the ground state. Mass of the electron = 9.1x10-31 kg and charge of the electron = 1.6x10-19 C.

Answer 1

We know that the Relation of Coulomb rule is given by the equation,

      F = kq₁q₂/r²

Also,

Centripetal Force = mv²/r

Since, Both the Forces are equal,

∴ mv²/r = kq₁q₂/r²

⇒ v²/r = kq²/r²m

v² = kq²/rm

⇒ v² = 9 × 10⁹ × (1.6 × 10⁻¹⁹)²/5.3 × 10⁻¹¹ × 9.1 × 10⁻³¹

⇒ v²= 0.479 × 10²⁰ × 10⁻³⁸ × 10³¹

⇒ v² = 4.78 × 10¹²

∴ v = 2.16 × 10⁶ m/s.

Hence, the velocity of the electrons is 2.16 × 10⁶ m/s.

Hope it helps.