If $K_1$ and $K_2$ are maximum kinetic energies of photoelectrons emitted when lights of wavelengths $\lambda_1$ and $\lambda_2$, respectively incident on a metallic surface and $\lambda_1=3 \lambda_2$, then

According to Einstein's photoelectric equation, the maximum kinetic energy $K$ of emitted photoelectrons is given by $K = \dfrac{hc}{\lambda} - \phi$, where $\phi$ is the work function of the metal.

For the first wavelength $\lambda_1$, $K_1 = \dfrac{hc}{\lambda_1} - \phi$.

For the second wavelength $\lambda_2$, $K_2 = \dfrac{hc}{\lambda_2} - \phi$.

Given $\lambda_1 = 3\lambda_2$, we substitute this into the expression for $K_1$:

$K_1 = \dfrac{hc}{3\lambda_2} - \phi = \dfrac{1}{3} \left( \dfrac{hc}{\lambda_2} - 3\phi \right)$.

Since $K_2 = \dfrac{hc}{\lambda_2} - \phi$, we have $\dfrac{hc}{\lambda_2} = K_2 + \phi$.

Substituting this into the expression for $K_1$:

$K_1 = \dfrac{1}{3} (K_2 + \phi - 3\phi) = \dfrac{K_2 - 2\phi}{3} = \dfrac{K_2}{3} - \dfrac{2\phi}{3}$.

Since $\phi > 0$, it follows that $K_1 < \dfrac{K_2}{3}$.

Answer: $K_1 < \left(\dfrac{K_2}{3}\right)$