Question

In the Bohr’s Hydrogen model, the electron is imagined to move in a circular orbit about a stationary proton. The force responsible for the electron circular motion is the electric force between the electron and the proton. Given that the radius of the electron’s orbit is 5.29x10-11m, and its mass is me=9.11x10-31km. Find the electron’s speed.

Answer 1

If the electron moves on the circular orbit, the gravitational force is balanced by the centrifugal force. Therefore, we may write the equality $F_g = F_c$ or $k_e\cdot\dfrac{q_pq_e}{r^2} = m_e\cdot \dfrac{v^2}{r}$ , where $k_e$ is the constant in the Coulomb's law, $q_p, q_e$ are the charges of the proton and electron, respectively. We should determine the speed v, so we rearrange the terms in equation to obtain

$v = \sqrt{\dfrac{k_e}{m_e}\cdot\dfrac{q_pq_e}{r}} = \sqrt{\dfrac{9\cdot10^9\,\mathrm{kg⋅m^3⋅s^{−2}⋅C^{−2}}}{9.11\cdot10^{-31}\,\mathrm{kg}}\cdot\dfrac{(1.6\cdot10^{-19}\,\mathrm{C})^2}{5.29\cdot10^{-11}\,\mathrm{m}}} = 2.2\cdot10^6\,\mathrm{m/s}.$