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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)^(+)), two protons are separated by roughly 1.5 Å and the electron is roughly1 Åfrom each proton. Determinethe potential energyof the system. Specify your choice of the zero of potential energy. |
Answer» Solution :Consider zero potential energy at infinity.  Here, CHARGE on electron, `q_(e)=-1.6xx10^(-19)C` Charge on proton, `q_(p)=1.6xx10^(-19)C` Distance betweenelectron and proton, `r_(ep)=1Å=1xx10^(-10)m` Distance between two protons, `r_(p p)=1.5Å =1.5xx 10^(-10)m` Then, electric POTENTIALENERGY of the system can be CALCULATED as: `U=(1)/(4pi epsilon_(0)) ((q_(e)q_(p))/(r_(ep))+(q_(e)q_(p))/(r_(ep))+(q_(p)q_(p))/(r_(p p)))` `rArr U=9xx10^(9) (((-1.6xx10^(-19))(1.6xx10^(-19)))/(1xx10^(-10))+((-1.6xx10^(-19))(1.6xx10^(-19)))/(1xx10^(-10))+((1.6xx10^(-19))(1.6xx10^(-19)))/(1.5xx10^(-10)))` `rArr U=-30.72xx10^(-19)J` `rArr U=19.2eV` |
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