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 = 2 K _ 2 `
D. ` K _ 2 = 2 K _ 1 `

Answer

B

Explanation

`K_(1)=(hc)/(lamda_(1))-phi_(0)........(i)`
`and K_(2)=(hc)/(lambda_(2))-phi_(0)or (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))"From (ii)"`
`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)`