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. $(\sqrt{150} = 12.27)$

The work done to accelerate an electron from rest so that it can have a de Broglie wavelength of $6600 \text{ Å}$ is nearly (Planck's constant $= 6.6 \times 10^{-34} \text{ J s}$ and mass of electron $= 9 \times 10^{-31} \text{ kg}$)

The de Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is ...... .

The de-Broglie wavelength of an electron having $80 eV$ energy is nearly ($1 eV = 1.6 \times 10^{-19} J$,Mass of the electron $= 9 \times 10^{-31} kg$,Planck's constant $= 6.6 \times 10^{-34} J-s$). (in $Å$)

An electron,a doubly ionized helium ion $(He^{++})$ and a proton have the same kinetic energy. The relation between their respective de-Broglie wavelengths $\lambda_{e}, \lambda_{He^{++}}$ and $\lambda_{P}$ is:

The resolving power of an optical instrument depends on the wavelength of the light used. An electron microscope uses a beam of electrons. Its resolving power can be increased if we: