Consider the Hydrogen atom to be a proton embedded in cavity of radius (Bohr radius) whose charge is neutralised by addition of electron to the cavity in a vacuum infinitely slowly.Estimate the average total energy of an electron in its ground state as the work done in the above process.Also if the magnitude of the average Kinetic energy is half of magnitude of average potential energy find the average potential energy.

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Solution

Work done while bringing an electron infinitely slowly from infinity to proton of radius a₀ is given as follows
W = - e²/4πε₀.a₀
NOTE : This work done is equal to the total energy of an electron in its ground state in the hydrogen atom. At this stage, the electron is not moving and do not possess any K.E., so this total energy is equal to the potential energy.
T.E. = P.E + K.E. = e²/4πε₀.a₀ ….(i)
In order the electron to be captured by proton to form a ground state hydrogen atom it should also attain
K.E. = e²/8πε₀a₀
(it is given that magnitude of K.E. is half the magnitude of P.E. Note that P.E. is –ve and K.E is +ve)
∴ T.E. = P.E + K.E. = - e²/4πε₀a₀ + e²/8πε₀a₀
or T.E. = - e²/8πε₀a₀
P.E. = 2 * T.E. = 2 * -e²/8π ∈₀a₀ or P.E. = -e²/8π ∈₀a₀

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