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) Potential energy of the system of two charges is represented as:
`U=(1)/(4pi epsilon_(0)) (q_(1)q_(2))/(r )`
The potential energy of the given system is calculated as:
`U=(9xx10^(9)xx(-1.6xx10^(-19))xx1.6xx10^(-19))/(5.3xx10^(-11)) `
`=-43.47xx10^(-19)J`
`U=-(43.47xx10^(-19))/(1.6xx10^(-19))`
`=-27.16eV=27.2 eV` (approx.)
(b) As it is given that the kinetic energy of the electron in the given orbit is half the magnitude of the potential energy of the given system.
Thus, kinetic energy of the electron,
`K=(1)/(2)xx27.2eV=13.6eV`
Thus, total energy of electron
`=U+K=(-27.2)eV+13.6eV=-13.6eV`
Total energy of the electron when it becomes free (i.e. taken to infinity) = 0
Work required to free the electron
`=0-(-13.6)=13.6eV`
(c ) Potential energy of the system when the separation between the charges is 1.06 Å is given by:
`U_(1)=9xx10^(9) xx ((-1.6xx10^(-19))xx(1.6xx10^(-19)))/((1.06xx10^(-10)))`
`=-21.73 xx 10^(-19)J`
Or, `U_(1)=-13.58eV = -13.6 eV` (approx.)
If zero of potential energy is taken at separation of 1.06 Å then the potential energy of the system is calculated as:
`U.=U-U_(1)`
Thus, `U.=-27.2-(-13.6)=-13.6eV`
For part (b), the minimum work done to free the electron will remain the same, that is 13.6 eV, even if the reference point is taken somewhere else.