Ground state energy of H-atom is `(-E_(1))`,t he velocity of photoelectrons emitted when photon of energy `E_(2)` strikes stationary `Li^(2+)` ion in ground state will be:

Options

A. `v=sqrt((2(E_(2)-E_(1)))/(m))`
B. `v=sqrt((2(E_(2)+9E_(1)))/(m))`
C. `v=sqrt((2(E_(2)-9E_(1)))/(m))`
D. `v=sqrt((2(E_(2)-3E_(1)))/(m))`

Answer

C

Explanation

Threshold energy of `Li^(2+)=9E_(1)`
Absorbed energy=Threshold energy+Kinetic energy of photoelectrons
`E_(2)+9E_(1)+(1)/(2)mv^(2)`
`mv^(2)=2(E_(2)-9E_(1))`
`v=sqrt((2(E_(2)-9E_(1)))/(m))`.