Answer
The electron has a longer de Broglie wavelength.
Work Step by Step
Since the magnitude of the charge of a proton and an electron are equal, both particles have the same kinetic energy after being accelerated through the potential difference.
We can write an expression for the speed of each particle:
$K = \frac{1}{2}mv^2$
$v = \sqrt{\frac{2~K}{m}}$
We can write an expression for the de Broglie wavelength:
$\lambda = \frac{h}{mv}$
$\lambda = \frac{h}{(m)(\sqrt{\frac{2~K}{m}})}$
$\lambda = \frac{h}{\sqrt{2~K~m}}$
Since $m_e \lt m_p$, then $\lambda_e \gt \lambda_p$
The electron has a longer de Broglie wavelength.
