Use Einstein’s photoelectric equation

For incident light of wavelength $\lambda$, the maximum kinetic energy of emitted photoelectrons is

$K_{\max}=h\nu-\phi=\frac{hc}{\lambda}-\phi$

where $\phi$ is the work function of the metal.

Write expressions for $K_1$ and $K_2$

Given wavelengths $\lambda_1$ and $\lambda_2$:

$K_1=\frac{hc}{\lambda_1}-\phi$ $K_2=\frac{hc}{\lambda_2}-\phi$

Also given:

$\lambda_1=3\lambda_2$

Hence,

$\frac{hc}{\lambda_1}=\frac{hc}{3\lambda_2}=\frac{1}{3}\cdot \frac{hc}{\lambda_2}$

So,

$K_1=\frac{1}{3}\frac{hc}{\lambda_2}-\phi$

Compare $K_1$ with $\dfrac{K_2}{3}$

First compute:

$\frac{K_2}{3}=\frac{1}{3}\left(\frac{hc}{\lambda_2}-\phi\right)=\frac{1}{3}\frac{hc}{\lambda_2}-\frac{\phi}{3}$

Now compare with

$K_1=\frac{1}{3}\frac{hc}{\lambda_2}-\phi$

Since $\phi>0$ for a metal,

$-\phi< -\frac{\phi}{3}$

Therefore,

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

Check options

• A: $K_1>\dfrac{K_2}{3}$ — false
• B: $K_1<\dfrac{K_2}{3}$ — true
• C: $K_1=\dfrac{K_2}{3}$ — false
• D: $K_2=\dfrac{K_1}{3}$ — false

Final answer

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

So the correct option is B.