Consider the hydrogen atom to be a proton embededded in a cavity of radius (Bohr radius) whose charge is neutralised by the addition of an electron to the cavity in a vacume innitially slowly .Extimate the average total energy of an electron in to ground state .Also if the magnitude iof the average kinetic energy is half of the magnitude of teh average energy ,find the average potential energy

Explanation

The work done in the neutralisation process in the energy due to the electric forrce of atteraction the electron and the proton
Potential energy = `(lambda Z e xx e)/(a_(0)) = (lambda Ze^(2))/(a_(0)) = (-lambda e^(2))/(a_(0))`
where e is the charge on the electron `a_(0)` is the Bohr radius , k is the coulonmbic law constant and Z is the atomic number `= 1` ,WE know energy of light obserbed in one photon
`= (hc)/(lambda_("obserbed"))`
Let `n_(1)` photon be obserbed .Therefore
Total energy obserbed `= (n_(1)hc)/(lambda"obserbed" )`
Now energy of light re-emited then,
Total energy re-emited out `= n_(2) xx (hc)/(lambda_("obserbed"))`
As given `E_("Absorbed") xx (47)/(100) = E_("Re-emmited out" )`
`(lambda c )/(lambda_("Absorbed")) xx n_(1) xx (47)/(100) xx (lambda_("Emitted"))/(lambda_("Absorbed"))`
`(n_(2))/(n_(1)) = (47)/(100) xx (5080)/(4530) = (," one proton")`
KInetic energy of the electron moving in the orbit `= (1)/(2) mv^(2) ("m is the mass of electron v is the velocity")`
Total energy of the electron
`E = (-lambda e^(2))/(a_(0)) + (1)/(2) mv^(2)`
For the electron to stay stable in the orbit , the electrostatic force of attraction should be equal to the centripetal force so
`-(ke^(2))/(a_(0)^(2)) = (mv^(2))/(a_(0))`
`:. -(ke^(2))/(a_(0)) = mv^(2)`
Sobstitating in (i) , we get
`E = (-ke^(2))/(a_(0)) + (1)/(2) (ke^(2))/(a_(0)) = - (ke^(2))/(2a_(0))`