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 1.5Å, and the electron is roughly 1Å from each proton. Determine the potential energy of the system. Specific your choice of the zero of potential energy.

Explanation

`q_1=` Charge on electrogen `=1.6xx10^-19C`
`q_2.q_3`= Charge on two protons
`= 1.6xx10^-19C`
Distance between `q_1` and `q_2` is
`r_12 = 1A^@`
`=1xx10^-10m`
Distance between `q_2` and `q_3` is
`r_23 = 1.5A^@`
`=1.5xx10^-10m`
Distance between `q_3` and `q_1` is
`r_31 = 1A^@ = 10^-10m`
Potential energy
`P.E. = 1/(4piepsilon_0)[(q_1q_2)/r_12+(q_2q_3)/r_23+(q_3q_1)/r_31]`
`=9xx10^9[((-1.6cc10^-19)xx(1.6xx10^-19))/(1xx10^-10)+(1.6xx10^-19)^2/(1.5xx10^-10)+((1.6cc10^-19)xx(-1.6xx10^-19))/(1xx10^-10)]`
`=(9xx10^(+9))/10^-10xx10^-38xx[(-1.6)^2+(1.6)^2/5-(1.6)^2]`
`=9xx10^9xx10^-28xx3.42`
`P.E. = -30.78xx10^-19J`