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

15

Explanation

`v = v_(0) (Z^(2))/(n^(3)) implies (Z^(2))/(n^(3)) = 2 (i)`
`E = E_(0) (Z^(2))/(n^(2)) implies (Z^(2))/(n^(3)) = 4 (ii)`
Solving (i) and (ii), `n = 2, Z = 4`
`L = mvr = (nh)/(2 pi), Delta L = tau Delta r = Delta n (h)/(2 pi)`
`tau = (Delta n)/(Delta t) xx (h)/(2 pi) = (1)/(7 xx 10^(-9)) xx (1)/(2) xx 2.1 xx 10^(-34) = (2.1)/(14) xx 10^(-25)`
`tau = 10^(27) = (2.1)/(14) xx 10^(-25) xx 10^(27) = 15`