An electron microscope uses electrons of $40 \ keV$. The de Broglie wavelength associated with these electrons is approximately:

The velocity of a particle $A$ is $3$ times the velocity of a proton. If the ratio of the de Broglie wavelengths of the particle $A$ and the proton is $3:2$,the mass of the particle $A$ is (where $m_{p}$ is the mass of the proton).

If the momentum of an electron is changed by $\Delta p,$ then the de-Broglie wavelength associated with it changes by $0.50\%.$ The initial momentum of the electron will be:

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 $nm$)

The ratio of de-Broglie wavelength of molecules of hydrogen and helium in two jars kept separately at temperatures of $27\,^{\circ}\text{C}$ and $127\,^{\circ}\text{C}$ respectively is

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are $\lambda_1$ and $\lambda_2$,their de Broglie wavelength in the frame of reference attached to their centre of mass is