In a hydrogen-like atom , an electron is orbating in an orbit having quantum number `n`. Its frequency of revolution is found to be `13.2 xx 10^(15)` Hz. Energy required to move this electron from the atom to the above orbit is `54.4 eV`. In a time of ? nano second the electron jumps back to orbit having quantum number`n//2`.If `tau` be the average torque acted on the electron during the above process, then find `tau xx 10^(27)` in Nm. (given: `h//lambda = 2.1 xx 10^(-34) J-s`, friquency of revolution of electron in the ground state of `H` atom `v_(0) = 6.6 xx 10^(15)` and ionization energy of `H` atom `(E_(0) = 13.6 eV)`

Answer

3

Explanation

`f=f_(0) (z^(2))/(n^(3))implies(z^(3))/(n^(3))=2`………i
`E=E_(0)(z^(2))/(n^(2))implies(z^(2))/(n^(2))=4`……….ii
Solving i and ii `n=2 z=4`
`L=mvr=(nh)/(2pi)`
`/_\L=mvr=(nh)/(2pi)`
`/_\L=tau/_\t=/_\(nh)/(2pi)`
`tau=(/_\n)/(/_\t).h/(2pi)`
`tauxx10^(27)=15`