Question

The average separation between the proton and the electron in a hydrogen atom in ground state is $5.3\times {10}^{-11m}$ (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?

Answer 1

Let $F_e$ and $F_G$ are electric force and gravitational force between two protons

$\therefore \frac{F_e}{F_g}=\frac{\frac{9\times {10}^9\times (1.6{)}^2\times {10}^{-38}}{r^2}}{\frac{6.67\times {10}^{-11}\times (1.732{)}^2\times {10}^{-54}}{r^2}}$

$=\frac{9\times (1.6{)}^2\times {10}^{-29}}{6.67\times (1.732{)}^2\times {10}^{-65}}$

$=1.24\times {10}^{36}$

Average separation between proton and electron of Hydrogen atom,

$r=5.3\times {10}^{-11}m$

(a) Coulomb's force,

$F=9\times {10}^9\times \frac{q_1{q}_2}{r^2}$

$=\frac{9\times {10}^9\times (1.0\times {10}^{-19}{)}^2}{(5.3\times {10}^{-11}{)}^2}$

(b) When the average distance between proton and electron becomes 4 times that of its ground state, then Coulomb's force,

$eF=\frac{1}{4\pi {\in }_0}\frac{q_1{q}_2}{(4r{)}^2}$

$=\frac{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}{16\times (5.3{)}^2\times {10}^{-22}}$

$=\frac{9\times (1.6{)}^2}{10\times (5.3{)}^2}\times {10}^{-7}$

$=0.0512\times {10}^{-7}$

$=5.1\times {10}^{-9}N$