Let  $K_1$  be the maximum kinetic energy of photoelectrons emitted by a light of wavelength ${\lambda }_1$   and $K_2$  corresponding to ${\lambda }_2$ .  If ${\lambda }_1=3{\lambda }_2$ , then

According to Photoelectric equation 

 $KE=\frac{hc}{\lambda }-\phi$

$\Rightarrow K_1=\frac{hc}{{\lambda }_1}-\phi$...........(1)

$\Rightarrow K_2=\frac{hc}{{\lambda }_2}-\phi =\frac{3hc}{{\lambda }_1}-\phi$.............(2)

Multiplying equation (1) by 3 and subtracting from equation (2) gives

$K_2-3K_1=2\phi$

$\Rightarrow K_1<\frac{K_2}{3}$