Question

If a hydrogen atom emit a photon of energy `12.1 eV` , its orbital angular momentum changes by `Delta L. then`Delta L` equals
A. `1.05xx10^(-34)`J-s
B. `211xx10^(-34)`J-s
C. `3.16xx10^(-34)`J-s
D. `4.22xx10^(-34)`J-s

Answer 1

Correct Answer - B
`DeltaE=-13.6eV(1/n_(1)^(2)-1/n_(2)^(2)), DeltaL=h/(2pi)(n_(1)-n_(2)`