Use Einstein’s photoelectric equation

For the fastest emitted electron, $K_{ ext{max}} = h\nu - \phi$ where $\phi$ is the work function.

Given in the question that the kinetic energy is very high compared to the work function, we take $K_{\text{max}} \approx h\nu = \frac{hc}{\lambda}$

Relate kinetic energy to momentum

For a non-relativistic electron, $K = \frac{p^2}{2m}$

Hence, $\frac{p^2}{2m} \propto \frac{1}{\lambda}$

So, $p^2 \propto \frac{1}{\lambda}$ and therefore $p \propto \frac{1}{\sqrt{\lambda}}$

Compare two cases

Let the initial wavelength be $\lambda$ and corresponding momentum be $p$.

For the new case, momentum is $1.5p = \frac{3p}{2}$.

Using $p^2 \propto \frac{1}{\lambda}$

we get $\frac{p_2^2}{p_1^2} = \frac{\lambda_1}{\lambda_2}$

Substituting $p_2 = 1.5p_1$, $\left(\frac{p_2}{p_1}\right)^2 = \frac{\lambda_1}{\lambda_2}$ $\left(\frac{3}{2}\right)^2 = \frac{\lambda}{\lambda_2}$ $\frac{9}{4} = \frac{\lambda}{\lambda_2}$

Therefore, $\lambda_2 = \frac{4}{9}\lambda$

Match with options

$\boxed{\lambda_2 = \frac{4}{9}\lambda}$

So the correct option is C.