The average separation between the proton and the electron in a hydrogen atom in ground state is `5.3xx10^-11m. a. Calculate the Coulomb force between tehm at this separation. b. When the atom goes into its first excited state the average separation between the roton and the electron increases to four times its value in the ground state.What is th Coulomb force in this state?

Answer

A, B

Explanation

Let `F_e and F_G` are electric force and gravitatioN/Al force between two protons
`:. F_e/F_g= (9xx10^9xx(1.6)^2xx10^-38)/r^2/(6.67xx10^011xx(1.732)^2xx10^-54)/r^2`
`= (9xx(106)^2xx10^-29)/(6.67xx(1.732)^2xx10^-65)`
` 1.24xx10^36`
Average separationbetween proton and electron of Hydrogn atom
`r=5.3xx10^-11m`
a. Coulomb's force,
`F=9xx10^9xx(q_1q_2)/r^2`
(9xx10^9xx(1.0xx10^-19)^2)/(5.3xx10^-11)^2)`
b. When the average distance between proton and electron becomes 4 times that of its ground state, then
Coulomb's force
`F= 1/(4pi epsilon_0) (q_1 q_2)/((4)^2)`
`=(9xx10^9xx(1.6xx10^-19)^2)/(16xx(5.3)^2xx10^-22)`
` (9xx(1.6)^2)/(10xx(5.3)^2) xx10^-7`
`=0.0512xx10^-7`
`=5.1xx10^-9N`