If a proton had a radius R and the chrge was uniformaly distributed, the ground state energy (ineV) of a H -atom for `R=0.1Å` is

Options

A. `-13.6`
B. `-27.2`
C. `-3.4`
D. `-30.8`

Answer

A

Explanation

In a H-atom in ground state, electron revolves round the point -size proton in a circular orbit of radius `r_B` (Bohr's radius).
As `mvr_(B)=ħ` and `(mv)^2/(r_(B))= (-1xxexxe)/(4piepsilon_0r_(B)^(2))`
`(m)/(r_B)((-ħ)/(m^2r_(B)^2))=(e^2)/(4piepsilon_0r_B^2),r_B=(4piepsilon_0)/(e^2)(ħ^2)/(m)=0.53Å`
`K.E. =(1)/(2)mv^2=((m)/(2))((ħ)/(mr_B))=(ħ^2)/(2mr_B^2)=13.6eV`
PE of the electron and proton,
`U=(1)/(4piepsilon_0)(e(-e))/(r_B)=-(e^2)/(4piepsilon_0r_B)=-27.2eV`
Total energy of the electron,i.e.,
`E=K.E.+U=+13.6eV-27.2eV=-13.6eV`
When `R=0.1Å:Rltr_(B)`(as `r_B=0.51Å`)and the ground stae energy is the same as obtained earlier for point size proton i.e ., `-13.6eV`.