An electron and a proton are accelerated through an identical...

The de Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by $\lambda = \dfrac{h}{\sqrt{2mqV}}$.

For an electron and a proton, the magnitude of the charge $q$ and the accelerating potential difference $V$ are identical.

This implies that $\lambda \propto \dfrac{1}{\sqrt{m}}$.

Since the mass of an electron is much less than the mass of a proton ($m_e < m_p$), the de Broglie wavelength of the electron will be greater than that of the proton.

Therefore, $\lambda_e > \lambda_p$.