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InterviewSolution
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In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å: (a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton. (b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)? (c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation? |
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Answer» The distance between electron-proton of a hydrogen atom, `d = 0.53 Å` Charge on an electron, `q_(1) = -1.6 xx 10^(-19)C` Charge on proton, `q_(2) =+1.6 xx 10^(-19)C` (a) Potential at infinity is zero. Potential energy of the system, p-e = Potential energy ay infinity - Potential energy at `= 0-(q_(1)q_(2))/(4pi in_(0)d)` Where, `in_(0)` is the permittivity of three space `(1)/(4pi in_(0)) = 9 xx 10^(9) Nm^(2) C^(-2)` `:.` Potential energy `= 0 -(9 xx 10^(9) xx (1.6 xx 10^(-19))^(2))/(0.53 xx 10^(10)) = -43.7 xx 10^(-19)J` Since `1.6 xx 10^(-19)J = 1eV`. `:.` Potential energy `= (-43.7 xx 10^(-19))/(1.6 xx 10^(-19)) = -27.2 eV` Therefore, the potential energy of the system is `-27.2 eV`. (b) Kinetic energy is half ofthe magnitude of potential energy, Kinetic energy `= (1)/(2) xx (-27.2) = 13.6 eV` Total energy `= 13.6 - 27.2 =13.6 eV` Therefore, the minimum work required to free the electron is `13.6 eV`. (c) When zero of potential energy s taken, `d_(1) = 1.06 Å` `:.` Potential energy of the system = Potential energy at `d_(1)` - Potential energy at d `= (q_(1)q_(2))/(4pi in_(0)d_(1)) - 27.2 eV` `= (9 xx 10^(9) xx(1.6 xx 10^(-19))^(2))/(1.06 xx 10^(-10)) - 27.2 eV` `= 21.73 xx 10^(-19) J -27.2 eV` `= 13.58 eV - 27.2 eV` `= -13.6 eV` |
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