An electron and a proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is

If the de Broglie wavelength of an electron is $2 \ nm$,then its kinetic energy is nearly (Planck's constant $= 6.6 \times 10^{-34} \ J \ s$ and mass of electron $= 9 \times 10^{-31} \ kg$) (in $eV$)

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 ratio of de Broglie wavelengths of a proton and an $\alpha$-particle having the same energy is ............

The kinetic energy of an electron with a de-Broglie wavelength of $0.3 \, nm$ is ............. $eV$.

$(a)$ Obtain the de Broglie wavelength of a neutron of kinetic energy $150 \; eV$. An electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. $(m_{n} = 1.675 \times 10^{-27} \; kg)$
$(b)$ Obtain the de Broglie wavelength associated with thermal neutrons at room temperature $(27 \; ^\circ C)$. Hence,explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.