Question

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?

Answer 1

The average separation between proton and electron of Hydrogen is $r=5.3{10}^{-11}m$.
when the average distance between proton and electron becomes $4$ times that of its ground state
Coulomb's force $F=\dfrac{1}{4\pi \epsilon_0}
\times \dfrac{q_1q_2}{(4r)^2}=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{16\times (5.3)^2\times 10^{-22}}=\dfrac{9\times (1.6)^2}{16\times (5.3)^2}\times 10^{-7}$
$=0.0512\times {10}^{-7}=5.1\times {10}^{-9}N$