The average separation between the proton and the electron in a hydrogen atom in ground state is 5.3 × 10^{−11} m. (a) Calculate the Coulomb force between them at this separation. (b) When the atom goes into its first excited state the average separation between the proton and the electron increases to four times its value in the ground state. What is the Coulomb force in this state?

#### Solution

Average separation between the proton and the electron of a Hydrogen atom in ground state, r = 5.3 × 10^{−11} m

(a) Coulomb force when the proton and the electron in a hydrogen atom in ground state

\[F = 9 \times {10}^9 \times \frac{q_1 q_2}{r_2}\]

\[ = \frac{9 \times {10}^9 \times \left( 1 . 6 \times {10}^{- 19} \right)^2}{\left( 5 . 3 \times {10}^{- 11} \right)^2} = 8 . 2 \times {10}^{- 8} \text{N}\]

(b) Coulomb force when the average distance between the proton and the electron becomes 4 times that of its ground state

\[\text {Coulomb force, F } = \frac{1}{4\pi \in_0} = \frac{q_1 q_2}{\left( 4r \right)^2}\]

\[ = \frac{9 \times {10}^9 \times \left( 1 . 6 \times {10}^{- 19} \right)^2}{16 \times \left( 5 . 3 \right)^2 \times {10}^{- 22}}\]

\[ = \frac{9 \times \left( 1 . 6 \right)^2}{10 \times \left( 5 . 3 \right)^2} \times {10}^{- 7} \]

\[ = 0 . 0512 \times {10}^{- 7} \]

\[ = 5 . 1 \times {10}^{- 9} \text{ N }\]