In the hydrogen atom, the energies corresponding to first, second and third orbits are given by `E_1` = -13.6 eV, `E_2` = -3.5eV, `E_3` = -1.5 eV.
If the atom emits a photon of energy 12.1 eV, in a transition, then the corresponding change in the angular momentum is given by

Options

A. `(h)/(2pi)`
B. `h/pi`
C. `(3h)/(2pi)`
D. `(2h)/(pi)`

Answer

B

Explanation

In this case , the transition is from n = 3 to n=1
[-1.5+13.6=12.1]
and `L_3 = (3h)/(2pi) " and " L_1 = (h)/(2pi)`
`therefore Delta L = L_3 - L_1 = (h)/(pi)`