An electron,an $\alpha$-particle,and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?

The de-Broglie wavelength associated with an electron,accelerated through a potential difference of $121 \ V$ is about:
[Take Planck's constant $h = 6.6 \times 10^{-34} \ J \cdot s$,mass of electron $m = 9 \times 10^{-31} \ kg$,charge of electron $e = 1.6 \times 10^{-19} \ C$] (in $nm$)

An electron (mass $m$) with initial velocity $\overrightarrow{v} = v_{0} \hat{i} + v_{0} \hat{j}$ is in an electric field $\overrightarrow{E} = -E_{0} \hat{k}$. If $\lambda_{0}$ is the initial de-Broglie wavelength of the electron,its de-Broglie wavelength at time $t$ is given by:

The additional energy that should be given to an electron to reduce its de-Broglie wavelength from $1 \ nm$ to $0.5 \ nm$ is

If a proton and an alpha particle are accelerated through the same potential difference, then the ratio of their de Broglie wavelengths is

If particles are moving with the same velocity,then the maximum de-Broglie wavelength will be for