If one of the two electrons of a `H_2` molecule id removed , we get a hydrogen molecules ion `H_(2)^(+)`. In the ground state of an `H_(2)^(+)`. The two protons are separated by roughly `1.5Å`. And the electron is at 1Å roughly from the nucleus. Find potential energy of the system. Specigy your choice of the zero of potential energy.

Explanation

Here `q_(1)` = charge on electron
`(= - 1.6 xx 10^(-19)C)`
`q_(2), q_(3)` = charge on two protons,
each `= 1.6 xx 10^(-19)C`
`r_(12)` = distance between
`q_(1)` and `q_(2) = 1 Å = 10^(-10) m`
`r_(23)` = distance between
`q_(2)` and `q_(3) = 1.5 Å = 1.5 xx 10^(-10) m`
`r_(31)` = distance between
`q_(3)` and `q_(1) = 1 Å = 10^(-10) m`.
Taking zero of potential energy at infinity we have
`P.E = (1)/(4 pi epsilon_(0)) [(q_(1) q_(2))/(r_(12)) + (q_(2) q_(3))/(r_(23)) + (q_(3)q_(1))/(r_(31))]`
`= 9 xx 10^(9) [((-1.6 xx 10^(-19)) xx (1.6 xx 10^(-19)))/(10^(-10)) + ((1.6 xx 10^(-19))^(2))/(1.5 xx 10^(10)) + ((1.6 xx 10^(-19))(-1.6 xx 10^(-19)))/(1 xx 10^(-10))]`
`= (9 xx 10^(9))/(10^(-10)) xx 10^(-38) xx [-1.6 xx 1.6 + ((1.6)^(2))/(1.5) - (1.6)^(2)]` ltbr. `= - 9 xx 10^(-19) xx 3.42 = - 30.78 xx 10^(-19) J`.