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

Options

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

Answer

B

Explanation

Here work function, `phi_0 = 0.5 eV`
According to Einstein.s photoelectric equation
Maximum kinetic = Incident -- Work function
energy of the emitted `" "` photon
electrons `" "` energy
`:. K_(max_1)= 1 eV - 0.5 eV = 0.5 eV" "...(i)`
and `K_(max_2)= 2.5 eV - 0.5 eV = 2 eV" "...(ii)`
Divide (i) by (ii), we get
`(K_("max"_1))/(K_("max"_2))=(0.5 eV)/(2eV) =1/4`
`(1/2mv_("max"_1)^2)/(1/2mv_("max"_2)^2)=1/4 " or " (v_("max"_1))/(v_("max"_2))=sqrt(1/4)=1/2`