The electron microscope is based on the principle of

The de-Broglie wavelength of a proton (charge $= 1.6 \times 10^{-19} \ C$,mass $= 1.67 \times 10^{-27} \ kg$) accelerated through a potential difference of $1 \ kV$ 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 $Å$)

$A$ nucleus $A$,with a finite de Broglie wavelength $\lambda_A$,undergoes spontaneous fission into two nuclei $B$ and $C$ of equal mass. $B$ flies in the same direction as that of $A$,while $C$ flies in the opposite direction with a velocity equal to half of that of $B$. The de Broglie wavelengths $\lambda_B$ and $\lambda_C$ of $B$ and $C$ are respectively

What is a wave packet?

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