Two photons of energies twice and thrice the work function of a metal are incident on the metal surface .Then, the ratio of maximum velocities of the photoelectrons emitted in the two cases respectively ,is

Options

A. `sqrt 2:1`
B. `sqrt3:3`
C. `sqrt3:sqrt2`
D. `1:sqrt2`

Answer

D

Explanation

Energy of first photon `E_(1)=2W`
Energy of second photon `E_(2)=3W`
The expression for `K.E _(max)` of photoelectron is given as , `K.E_(max)=E-W`
`implies (KE_(1))_(max)` of first photon -2W-W=W
`implies(K.E_(2))_(max)` of second photon =3W-W=2W
the ratio of maximum velocites can be calculated as ,
`((KE_(1))max)/((K.E_(2))max)=(1)/(2)`
`implies ((1)/(2)mv_(1)^(2))/((1)/(2)mv_(1)^(2))=(1)/(2)(therefore K.E=(1)/(2)mv^(2))``implies (v_(1))/(v_(2))=sqrt(1)/(2)`
`thereforev_(1):v_(2)=1:sqrt2`