The de-Broglie wavelength of electron
$$
\lambda_e=\frac{h}{\sqrt{2 m K_e}}
$$
$K_e=$ Kinetic energy of electron
$$
\lambda_e^2=\frac{h^2}{2 m_e K_e} \Rightarrow K_e=\frac{h^2}{\lambda_e^2 \cdot 2 m_e}
$$
De-Broglie wavelength of photon is $\lambda_p$.
Kinetic energy of photon
$$
\begin{aligned}
& K_P=h v=\frac{h c}{\lambda_P} \\
& \therefore \quad \frac{K_e}{K_P}=\frac{h^2 / \lambda_e^2 \cdot 2 m_e}{h c / \lambda_P}
\end{aligned}
$$
As, $\lambda_e=\lambda_p$
$$
\begin{aligned}
& \text { Then, } \frac{K_e}{K_P}=\frac{h}{c \lambda_e \times 2 m_e} \\
& =\frac{6.66 \times 10^{-34}}{3 \times 10^8 \times 1.2 \times 10^{-10} \times 2 \times 9.1 \times 10^{-31}} \cong \frac{1}{100} \\
& \therefore \quad K_e: K_P:: 1: 100
\end{aligned}
$$