If a hydrogen atom emit a photon of energy `12.1 eV` , its orbital angular momentum changes by `Delta L. then`Delta L` equals

Options

A. `1.05xx10^(-34)J" "sec`
B. `2.11xx10^(-34)J" "sec`
C. `3.16xx10^(-34)J" "sec`
D. `4.22xx10^(-34)J" "sec`

Answer

B

Explanation

Emission of photo of 12.1 eV corresponds to the `therefore` Change in angular momantum
`=(n_(2)-n_(1))(h)/(2pi)`
`=(3-1)(h)/(2pi)=(h)/(pi)`
`=(6.626xx10^(-34))/(3.14)=2.11xx10^(-34)J" "sec`