If the kinetic energy of a free electron doubles,its de-Broglie wavelength changes by the factor

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

Crystal diffraction experiments can be performed using $X-$rays,or electrons accelerated through an appropriate voltage. Which probe has greater energy? (For quantitative comparison,take the wavelength of the probe equal to $1\,\mathring{A}$,which is of the order of inter-atomic spacing in the lattice) $(m_{e}=9.11 \times 10^{-31}\,kg)$

The kinetic energies of an electron,$\alpha$-particle,and a proton are given as $4K, 2K$,and $K$ respectively. The de-Broglie wavelengths associated with the electron $(\lambda_e)$,$\alpha$-particle $(\lambda_\alpha)$,and the proton $(\lambda_p)$ are related as follows:

Consider two particles of different masses. In which of the following situations will the heavier of the two particles have a smaller de-Broglie wavelength?

The voltage applied to an electron microscope to produce electrons of wavelength $0.50 \text{ Å}$ is (in $\text{ V}$)