Given, change in kinetic energy of photoelectrons,
$$
\begin{aligned}
\Delta K & =0.52 \mathrm{eV} \\
\lambda_1 & =500 \mathrm{~nm}, \lambda_2=?
\end{aligned}
$$
We know that, kinetic energy of emitted photoelectron (using Einstein's equation)
$$
\begin{gathered}
K=\frac{h c}{\lambda}-\phi \\
\therefore \text { At wavelength } \lambda_1 \\
K_1=\frac{h c}{\lambda_1}-\phi
\end{gathered}
$$
At wavelength $\lambda_{2^{\prime}}$
$$
K_2=\frac{h c}{\lambda_2}-\phi
$$
From Eqs. (i) and (ii), we have
$$
\begin{array}{cc}
& K_2-K_1=\frac{h c}{\lambda_2}-\frac{h c}{\lambda_1} \\
\Rightarrow \quad \Delta K=h c\left(\frac{1}{\lambda_2}-\frac{1}{\lambda_1}\right) \\
\Rightarrow \quad 0.52=1242\left(\frac{1}{\lambda_2}-\frac{1}{500}\right)
\end{array}
$$

$$
\begin{array}{rlrl}
\Rightarrow & & \frac{0.52}{1242} & =\frac{1}{\lambda_2}-\frac{1}{500} \\
\Rightarrow & \frac{1}{\lambda_2} & =\frac{0.52}{1242}+\frac{1}{500} \\
\Rightarrow & & \lambda_2 & \approx 400 \mathrm{~nm}
\end{array}
$$