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

Answer

A

Explanation

According to Einstein's theory of photoelectric effect, we have energy of incidence photon
=KE of photoelectron+work function
i.e., `hv=(1)/(2)mv_(max)^(2)+W_(0)`
Given, `E_(1)=1eV` ltBrgt `E_(2)=2.5eV,phi_(0)=0.5eV`
We know that, `E=phi_(0)+K_(max)`
In the first condition, `1=0.5+K_(1)`
`K_(1)=0.5`
Similarly, `2.5=0.5+K_(2)`
`K_(2)=2` . . . (ii)
We know that `(v_(1))/(v_(2))=sqrt((K_(1))/(K_(2)))`
From eqs. (i) and (ii), we get `(v_(1))/(v_(2))=sqrt((1)/(4))`
`(v_(1))/(v_(2))=(1)/(2)`.