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

(b) : According to Einstein's photoelectric equation
`(hc)/(lamda)=phi_(0)+KE_(max)`
`K_(1)=(hc)/(lamda_(1))-phi_(0) and K_(2)=(hc)/(lamda_(2))-phi_(0)`
or `K_(1)-K_(2)=hc[(1)/(lamda_(1))-(1)/(lamda_(2))]`
`=hc[(1)/(3lamda_(2))-(1)/(lamda_(2))]=-(2hc)/(3lamda_(2))` `("Given"lamda_(1)=3lamda_(2))`
`K_(1)-K_(2)=-(2)/(3)(K_(2)+phi_(0))`
or `K_(1)=K_(2)-(2)/(3)K_(2)-(2)/(3)phi_(0)=(K_(2))/(3)-(2)/(3)phi_(0) or K_(1)lt(K_(2))/(3)`