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

Options

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

Answer

D

Explanation

Let `phi_0` be the work function of a metal surface .
Given : `E_1 = 2phi_(0) and E_(2) = 3phi_0`
According to Einstein. s photoelectric equation
Incident energy = work function + Maximum kinetic energy of an emitted electron
`:. E_(1) = phi_(0) +K_(max1) " "...(i) `
and ` E_(2) = phi_(0) +K_(max2) " "...(ii) `
or `K_(max1)=E_1 - phi_0 " "...(iii)`
or and `K_(max2)=E_2 - phi_0 " "...(iv )`
Divide (iii) by (iv) , we get `(K_(max1))/(K_(max2))=(E_1-phi_0)/(E_2-phi_0)`
Substituting the given values , we get
`(K_(max1))/(K_(max2))=((2phi_0-phi_0)/(3phi_0-phi_0))=1/2" or " (v_(max1)^2)/(v_(max2)^2)=1/2" or " (v_(max1))/(v_(max2))=1/sqrt2`