Answer to Question #108422 in Atomic and Nuclear Physics for Jonathan

Question

In a hydrogen atom, the proton and the electron are separated by an average distance 5 ∙ 10"## �. (a) Calculate the force that the proton exerts on the electron at this distance. (b) Calculate the electric field strength at the average location of the electron.

Answer 1

charge on electron and proton are of equal magnitude and of opposite sign and magnitude of charge =1.6x10^-19

Since I am not provided with the distance between the charges so I am assuming it as 'k'

Force between two charges is given by $k\frac{q_1q_2}{r^2}=9\times10^9\times\frac{1.6\times10^{-19}\times1.6\times10^{-19}}{k^2}=\frac{23.04\times10^{-29}}{k^2}N$

As the charges of opposite magnitude hence the force will be attractive in nature

In the next part we have to find electric field at the location of the electron

Electric field is given by $(k\frac{q_1}{r_1^2}+k\frac{q_2}{r_2^2})$ so if we see that as the distance of electron from its average location will be almost zero hence the electric field will become infinity