Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminate a metallic surface whose work function is 0.5 eV. Ratio of maximum speeds of emitted electrons will be

Options

A. `1:4`
B. `1:1`
C. `1:5`
D. `1:2`

Answer

D

Explanation

According to Einstein's photogenic equation , the maximum kinetic energy of emitted photoelectrons is
`K_(max) = h v - phi_(0)`
where hv is the energy of incident and `phi_(0)` is the work function .
But `K_(max) = 1/2 mv_(max)^(2)`
` :. 1/2 mv_(max)^(2) = hvv - phi _(0)`
As per equation
` 1/2 mv_(max) = 1 eV - 0.5 eV = 0.5 eV ....(i)`
and `and 1/2 mv_(max_(2))^(2) = 2.5 eV - 0.5 eV = 2 eV " " ...(ii)`
Dividing eqn . (i) by eqn . (ii) , we get
`(v_(max_(1))^(2))/(v_(max_(2))^(2)) = (0.5 eV)/(2 eV) = 1/4`
` v_(max_(1))/(v_(max_(2))) = sqrt(1/4) = 1/2 `