If one of the two electrons of a `H_(2)` molecule is removed, we get a hydrogen molecular ion `H_(2)^(+)`.In the ground state of an `H_(2)^(+)` , the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy

Explanation

System of charges is shown in the figure

Charge on electron `q_(1)=- 1.6xx10^(-19)` C
Charge on proton `q_(2)=q_(3)= 1.6xx10^(-19)` C
Talcing potential zero at infinity distance then, potential energy of the system,
`U = U_(12)+U_(23) +U_(13)`
`= k[(q_(1)q_(2))/(r_(1))+(q_(2)q_(3))/(2)+(q_(1)q_(3))/(r_(3))]`
`=9xx10^(9) [((-16xx10^(-19))(1.6xx10^(-19)))/(10^(-10))+((1.6xx10^(-19))^(2))/(1.5xx10^(-10))+(-(1.6xx10^(-19))^(2))/(10^(-10))]`
`=(9xx10^(9)xx(1.6xx10^(19)xx1.6xx10^(-19))^(2))/(10^(-10))[-1+(1)/(1.5)-1]`

`=(9xx10^(9)xx(1.6xx10^(19)xx1.6xx10^(-19))^(2))/(10^(-10))[(-3+1)/(1.5)]`
`=23.04xx10^(-10)xx((-2)/(1.5))`J
`=(-23.04xx10^(-19)xx2)/(1.6xx10^(-19)xx1.5)eV[because1.6xx10^(-19)` J = 1 eV]
=-19.2 eV