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

(a) Here `phi_(1) =- 1.6xx10^(-19) C`
`phi_(2)= +1.6xx10^(-19) ` C
`r = 0.53 Å= 0.53 xx10^(-10) ` m
Potential energy of system of electron and proton ,
`U= (kq_(1)q_(2))/(r)=(9xx10^(9)xx(-1.6xx10^(19))xx(1.6xx10^(9)))/(0.53xx106(-10))`
`:. U =- 43.4717xx10^(19)` J
`:. =-(43.4717xx10^(-19))/(1.6xx10^(-19))eV`
`:. U =- 27.1699 ` eV
`:. U = -27.2` eV
(b) Kinetic energy of orbital electron
K.E (kinetic energy = `(1)/(2)xx` potential energy
`=-(1)/(2)xx (-27.2)`
K.E = 13.6 eV
and total energy of electron
= potential energy + kinetic energy
`=-27.2 +13.6`
`= -13.6eV`
Total energy of free electron is zero, hence minimum work required to free the electron,
`W = K_(i) -K_(f)`
`= 0-(-13.6)`
`:. ` W = 13.6 eV
( c) Since total energy of free electron be zero then potential energy of electron and proton sytem
`U=Kq_(1)q_(2)[(1)/(r_(1))-(1)/(r_(2))]`
`=9xx10^(9)xx(-1.6xx10^(-19))xx(1.6xx10^(-19))`
`[(1)/(0.53xx10^(-10))-(1)/(1.06)xx10^(10)]`
`= -(9xx2.56xx10^(-28))/(0.53xx10^(10))[1-(1)/(2)]`
`=-(9xx2.56xx10^(-28)xx1)/(0.53xx106^(-10)xx2xx1.6xx10^(-19))` eV
`=-13.6` eV
This indicates that an electron is taken from `0.53 Å` to `1.06 Å,` 13.6 eV energy is used and so potential energy increases from - 27.2 eV to - 13.6 eV. In this state kinetic energy must be zero. Because total energy is zero hence minimum energy required to free electron = 0 - (- 13.6) = 13.6 eV.