The maximum potential energy due to electrostatic repulsion between two hydrogen nuclei is nearly (radius of the nucleus = 1.1 fermi) `[(1)/(4pi epsi_(0))= 9 xx 10^(9) Nm^(2) C^(-2)]`

Options

A. 0.65 MeV
B. 2.09 MeV
C. 3.31 MeV
D. 0.92 MeV

Answer

A

Explanation

Potential energy due to two charges `=(kq_(1)q_(2))/(d)`
For hydrogen atom, `q_(1) = q_(2) = 1.6 xx 10^(-19)C`
`U = (9 xx 10^(9) xx (1.6 xx 10^(-19))^(2))/(2.2 xx 10^(-15)) " " [ :. D =2 epsi]`
`=10.47 xx 10^(-14)J =(10.47 xx 10^(-4))/(1.6 xx 10^(-19))eV`
`=6.5 xx 10^(5) eV = 0.65MeV`