Question

Two protons are situated at a distance of 100 fermi from each other. The potential energy of this system will be in ev​

Answer 1

Explanation:

We know potential energy U =k

r

q

1

q

2

U =

0.53×10

−10

(9×10

9

)×(1.6×10

−19

)(1.6×10

−19

)

J

(1 J=1.6×10

−19

eV)

∴ U=+27.2eV