Two radiation of photons energy 3eV and 4.5 eV successively illuminate a photosensitive metallic surface of work function 2.5 eV the ratio of the maximum speeds of the emitted electrons is :-

Options

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

Answer

B

Explanation

According to Einstein.s photoelectric equation `1/2 mv_(max)^2 = hupsilon-phi_0`
where `1/2 mv_(max)^2` is the maximum kinetic energy of the emitted electrons, `hupsilon` is the incident energy and `phi_0` is the work function of the metal .
`:. 1/2mv_(max1)^2=1eV -0.5 eV = 0.5 eV " "....(i)`
and `1/2mv_(max2)^2=1eV -2.5 eV = 0.5 eV=2eV " "....(ii)`
Divide (i) and (ii) , we get
`(v_(max1)^2)/(v_(max2)^2)=(0.5)/2implies (v_(max1))/(v_(max2))=sqrt((0.5)/2)=1/2`