If `K_(1) and K_(2)` are maximum kinetic energies of photoelectrons emitted when light of wavelength `lambda_(1) and lambda_(2)` respectively are incident on a metallic surface. If `lambda_(1)=3lambda_(2)` then

Options

A. `K_(1) gt (K_(2))/(3)`
B. `K_(1) lt (K_(2))/(3)`
C. `K_(1) = 3K_(2)`
D. `K_(2) = 3K_(1)`

Answer

B

Explanation

`K_(1) = (hc)/(lambda_(1)) - phi " "or (hc)/(lambda_(1)) = K_(1)+phi`
`K_(2)=(hc)/(lambda_(2))-phi " " or (hc)/(lambda_(2)) = K_(2)+phi`
`lambda_(1) = 3 lambda_(2), (hc)/(3lambda_(2)) = K_(1)+phi`
`((K_(2)+phi)/(3)) =K_(1)+phi`
`(K_(2))/(3) = K_(1)+(2phi)/(3):. (K_(2))/(3) gt K_(1)`