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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.

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Answer Text

Solution :
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`.

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