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

Maximum kinetic energies of photo electrons for light of wavelength `K_(1) = (hc)/lambda - f`,
Maximum kinetic energies of photo electrons for light of wavelength `K_(2) = (hc)/lambda_(2) -f`
Also given, `lambda_(1) = 3lambda_(2)`
Therefore
`K_(2)/K_(1) gt 3`