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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 |
Answer» Solution :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 |
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