If the potential difference used to accelerate electrons is doubled,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:

The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of $n$ is

For which of the following particles will it be most difficult to experimentally verify the de Broglie relationship?

Two particles of equal masses are moving with equal speeds at an angle $60^o$. The de-Broglie wavelength of these particles is $\lambda$. Find the de-Broglie wavelength of the particles in the frame of the centre of mass of the particles.

The de-Broglie wavelength of an electron is $0.4 \times 10^{-10} \ m$ when its kinetic energy is $1.0 \ keV$. Its wavelength will be $1.0 \times 10^{-10} \ m$ when its kinetic energy is: (in $keV$)