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

If a proton and an $\alpha$-particle have the same kinetic energy,what is the ratio of their de-Broglie wavelengths?

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:

An electron beam of energy $10 \, KeV$ is incident on a metallic foil. If the interatomic distance is $0.2454 \, \mathring{A}$,then find the angle of diffraction in degrees. $(\sqrt{150} = 12.27)$

$A$ bomb projected from the ground at an angle $\theta$ $\left( \theta \neq 90^\circ \right)$ explodes into two fragments of equal mass at the topmost point of its trajectory. If one of the fragments returns to the point of projection,then the ratio of the de Broglie wavelength of the second fragment just after the explosion to that of the bomb just before the explosion is:

The de Broglie wavelength of an electron having kinetic energy $E$ is $\lambda$. If the kinetic energy of the electron becomes $\frac{E}{4}$,then its de Broglie wavelength will be: