The ratio of de Broglie wavelength of molecules of hydrogen and helium in two gas jars kept separately at temperatures of $27\,^oC$ and $127\,^oC$ respectively is

The de Broglie wavelength $(\lambda)$ depends on mass '$m$' and kinetic energy '$E$' according to which formula?

What is the kinetic energy of a neutron associated with a de Broglie wavelength of $1.40 \times 10^{-10} \ m$? (Mass of neutron $m = 1.675 \times 10^{-27} \ kg$,$h = 6.6 \times 10^{-34} \ J \cdot s$).

After absorbing a slowly moving neutron of mass $m_{N}$ (momentum $0$),a nucleus of mass $M$ breaks into two nuclei of masses $m_{1}$ and $5m_{1}$ $(6m_{1} = M + m_{N})$ respectively. If the de-Broglie wavelength of the nucleus with mass $m_{1}$ is $\lambda$,then the de-Broglie wavelength of the other nucleus will be:

$A$ proton,an electron,and an alpha particle have the same kinetic energies. Their de-Broglie wavelengths will be compared as:

The de Broglie wavelength of an electron accelerated to a potential of $ 400 \,V $ is approximately (in $\,nm$)