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

Question #108422
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.
1
Expert's answer
2020-04-08T10:53:03-0400

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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS