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

The system of two protons and one electron is represented in the given figure. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_PHY_V01_XII_C02_E01_019_S01.png" width="80%"> <br> Charge on proton 1, `q_(1)=1.6xx10^(-19) C` <br> Charge on proton 2, `q_(2)=1.6xx10^(-19) C` <br> Charge on electron , `q_(3)=-1.6xx10^(-19)C` <br> Distance between protons 1 and 2 `d_(1)=1.5xx10^(-10)m` <br> Distance between proton 1 and electron `d_(2)=1xx10^(-10)` m <br> Distance between proton 2 and electron , `d_(3)=1xx10^(-10)m ` <br> the potential energy at infinity is zero . <br> potential energy of the system <br> `V=(q_(1)q_(2))/(4piin_(0)d_(1))+(q_(2)q_(3))/(4pi in_(0)d_(3)) +(q_(3)q_(1))/(4pi in_(0)d_(2))` <br> Substituting `1/(4pi in_(0))=9xx10^(9) Nm^(2) C^(-2)` , we obtain <br> `V=(9xx10^(9)xx10^(-19)xx10^(-19))/(10^(-19))[-(16)^(2)+((1.6)^(2))/1.5+(1.6)^(2)]` <br> `=-30.7xx10^(-19) J` <br> `=-19.2 eV` <br> Therefore, the potential energy of the system is `-19.2 eV`.