Question

Find the speed of the electron in the ground state of a hydrogen atom. The description of ground state is given in the previous problem.

Answer 1

Given:
Separation between the two charges, r = 0.53 Å = 0.53 × 10⁻¹⁰ m
By Coulomb's Law, force,     

\[F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}\]

Here, 

\[q_1  =  q_2  = e\]

\[\Rightarrow F = \frac{9 \times {10}^9 \times \left( 1 . 6 \times {10}^{- 19} \right)^2}{\left( 0 . 53 \times {10}^{- 10} \right)^2}\] 

\[     = 8 . 2 \times  {10}^{- 8}   N\]

Now, mass of an electron, M_e = 9.12 × 10⁻³¹ kg
The necessary centripetal force is provided by the Coulombian force.

\[\Rightarrow    F_e  = \frac{M_e v^2}{r}\] 

\[ \Rightarrow  v^2  = 0 . 4775 \times  {10}^{13} \] 

\[                     = 4 . 775 \times  {10}^{12}   \] 

\[ \Rightarrow v = 2 . 18 \times  {10}^6 \]  m/s