Question

A hydrogen atom in its ground state absorbs `10.2 eV` of energy. The orbital angular momentum is increased by
A. `1.05xx10^(-34) J-sec`
B. `2.11xx10^(-34) J-sec`
C. `3.16xx10^(-34) J-sec`
D. `4.22x10^(-34) J-sec`

Answer 1

Correct Answer - A
After absorbing 10.2 eV, electron goes to n=2 from n=1
`DeltaL=(n_(2)-n_(1))(h)/(2pi)=(2-1)(h)/(2pi)`
`=(6.6xx10^(-34))/(2xx3.14)=1.05xx10^(-34) J.sec`