If we consider an electron and a photon of the same de-Broglie wavelength,then they will have the same:

$A$ free particle with initial kinetic energy $E$ and de-Broglie wavelength $\lambda$ enters a region in which it has potential energy $V$. What is the particle's new de-Broglie wavelength?

The de Broglie wavelength for a deuteron can be given by:

An electron with mass $m$ and an initial velocity $(t=0)$ $\vec{v} = v_0 \hat{i}$ $(v_0 > 0)$ enters a magnetic field $\vec{B} = B_0 \hat{j}$. If the initial de Broglie wavelength at $t=0$ is $\lambda_0$,then its value after time $t$ would be:

$A$ proton and an $\alpha$-particle are accelerated using the same potential difference. How are the de-Broglie wavelengths $\lambda_p$ and $\lambda_{\alpha}$ related to each other?

$A$ proton and an electron are accelerated through the same potential difference. The ratio $\frac{\lambda_e}{\lambda_p}$ will be: