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))=(K_(2))/(3)`
D. `K_(2)=(K_(1))/(3)`

Answer

B

Explanation

`K_(1)=(hc)/(lambda_(1))-phi`
`K_(2)=(hc)/(lambda_(2))-phi`
`(K_(1))/(K_(2))=((hc)/(3lambda_(2))-phi)/((hc)/(lambda_(2))-phi)`
`K_(1)lt(K_(2))/(3)`