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?

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:

$A$ photon and an electron have equal energy $E$. The ratio of $\lambda(\text{electron})$ to $\lambda(\text{photon})$ is proportional to

The de Broglie wavelength $(\lambda)$ depends on mass '$m$' and kinetic energy '$E$' according to which formula?

The graph between the energy log $E$ of an electron and its de$-$Broglie wavelength log $\lambda$ will be

$A$ particle is travelling $4$ times as fast as an electron. Assuming the ratio of the de-Broglie wavelength of the particle to that of the electron is $2:1$,the mass of the particle is: