According to the de-Broglie hypothesis, the ratio of the wavelength of a photon and that of an electron having the same energy $E$ is $(m = \text{mass of electron}, c = \text{velocity of light})$

If the potential difference used to accelerate electrons is increased four times,by what factor does the de-Broglie wavelength associated with the electrons change?

If the de Broglie wavelengths of an electron and a proton are equal,then the kinetic energy of the electron is ........

The wavelength of light from the spectral emission line of sodium is $589 \; nm$. Find the kinetic energy at which
$(a)$ an electron,and
$(b)$ a neutron,would have the same de Broglie wavelength.

If the de Broglie wavelengths of an electron and a proton are equal,then what will be the kinetic energy of the electron?

Particle $A$ of mass $m_{A} = \frac{m}{2}$ moving along the $x$-axis with velocity $v_{0}$ collides elastically with another particle $B$ at rest having mass $m_{B} = \frac{m}{3}$. If both particles move along the $x$-axis after the collision,the change $\Delta \lambda$ in the de-Broglie wavelength of particle $A$,in terms of its de-Broglie wavelength $(\lambda_{0})$ before the collision is: