If `K_(1)` and `K_(2)` are maximum K.E. photoelectron emitted when lights wavelength `lambda_(1)` and `lambda_(2)` respectively incident a metallic surface .If `lambda_(1)=3lambda_(2)`,then…..

Options

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

Answer

B

Explanation

`K_(1)=(hc)/(lambda_(1))-phi_(1)` …..(i)and
`K_(2)=(hc)/(lambda_(2))-phi_(0)` and `(hc)/(lambda_(2))=(K_(2)+phi_(0))`…..(ii)
`therefore K_(1)-K_(2)=hc[(1)/(lambda_(1))-(1)/(lambda_(2))]`
`=hc[(1)/(3lambda_(2))-(1)/(lambda_(2))]=-(2hc)/(3lambda_(2))`
`=-(2)/(3)(K_(2)+phi_(0))[because` From result (1)]
`therefore K_(1)=K_(2)-(2)/(3)K_(2)-(2)/(3)phi_(0)=(K_(2))/(3)-(2)/(3)phi_(0)`
`therefore K_(1)lt(K_(2))/(3)`