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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.
Solution
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,
\[\Rightarrow F = \frac{9 \times {10}^9 \times \left( 1 . 6 \times {10}^{- 19} \right)^2}{\left( 0 . 53 \times {10}^{- 10} \right)^2}\]
\[ = 8 . 2 \times {10}^{- 8} N\]
Now, mass of an electron, Me = 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
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