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?

Explanation

Here `q_(1) = 1.6xx10^(-19) C , q_(2) = +1.6xx10^(-19) C, r = 0.53 Å = 0.53xx10^(-10) m`
Potential energy = P.E at `oo -` P.E at r
`= 0 - (q_(1) q_(2))/(4pi in_(0) r) = (-9xx10^(9) (1.6xx10^(-19))^(2))/(0.53xx10^(-10)) = -43.47xx10^(-19) joul e`
`= (-43.47xx10^(-19))/(1.6xx10^(-19)) eV = -27.16eV`
(b) K.E. in the orbit `= (1)/(2) (27.16)eV = 13.58eV`
Total energy ` = K.E. + P.E. = 13.58 - 27.16 = -13.58 eV`
work required to free the electron `= 13.58 eV`
(c ) Potental energy at a separation of `r_(1) (= 1.06 Å)` is
`= (q_(1) q_(2))/(4pi in_(0) r_(1)) = (9xx10^(9) (1.6xx10^(-19))^(2))/(1.06xx10^(-10)) = 21.73xx10^(-19) J = 13.58 eV`
Potential energy of the system, when zero of P.E. is taken at `r_(1) = 1.06 Å` is
`= P.E." at "r_(1) - P.E at r = 13.58-27.16 = -13.58 eV`.
By shifting the zero of potential energy, work required to free the electron is not affected. It continues to be the same, being equal to `+13.58 eV`.