In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å: <br> (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. <br> (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)? <br> (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 Text

Solution :
The distance between electron-proton of a hydrogen atom, `d = 0.53 Å` <br> Charge on an electron, `q_(1) = -1.6 xx 10^(-19)C` <br> Charge on proton, `q_(2) =+1.6 xx 10^(-19)C` <br> (a) Potential at infinity is zero. <br> Potential energy of the system, p-e = Potential energy ay infinity - Potential energy at `= 0-(q_(1)q_(2))/(4pi in_(0)d)` <br> Where, <br> `in_(0)` is the permittivity of three space <br> `(1)/(4pi in_(0)) = 9 xx 10^(9) Nm^(2) C^(-2)` <br> `:.` 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` <br> Since `1.6 xx 10^(-19)J = 1eV`. <br> `:.` Potential energy `= (-43.7 xx 10^(-19))/(1.6 xx 10^(-19)) = -27.2 eV` <br> Therefore, the potential energy of the system is `-27.2 eV`. <br> (b) Kinetic energy is half ofthe magnitude of potential energy, <br> Kinetic energy `= (1)/(2) xx (-27.2) = 13.6 eV` <br> Total energy `= 13.6 - 27.2 =13.6 eV` <br> Therefore, the minimum work required to free the electron is `13.6 eV`. <br> (c) When zero of potential energy s taken, `d_(1) = 1.06 Å` <br> `:.` Potential energy of the system = Potential energy at `d_(1)` - Potential energy at d <br> `= (q_(1)q_(2))/(4pi in_(0)d_(1)) - 27.2 eV` <br> `= (9 xx 10^(9) xx(1.6 xx 10^(-19))^(2))/(1.06 xx 10^(-10)) - 27.2 eV` <br> `= 21.73 xx 10^(-19) J -27.2 eV` <br> `= 13.58 eV - 27.2 eV` <br> `= -13.6 eV`

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