The figure shows four situations in which an electron is moving in an electric or magnetic field. In which case is the de Broglie wavelength of the electron increasing?

$A$ proton and an electron are associated with the same de-Broglie wavelength. The ratio of their kinetic energies is:
(Assume $h=6.63 \times 10^{-34} \ J \ s$,$m_{e}=9.0 \times 10^{-31} \ kg$ and $m_{p}=1836 \times m_{e}$)

Kinetic energy of a proton is equal to energy $E$ of a photon. Let $\lambda_1$ be the de-Broglie wavelength of the proton and $\lambda_2$ be the wavelength of the photon. If $\frac{\lambda_1}{\lambda_2} \propto E^{n}$,then the value of $n$ is

$(a)$ 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.

The de Broglie wavelength of an oxygen molecule at $27^{\circ} C$ is $x \times 10^{-12} \ m$. The value of $x$ is (take Planck's constant $= 6.63 \times 10^{-34} \ J \cdot s$, Boltzmann constant $= 1.38 \times 10^{-23} \ J/K$, mass of oxygen molecule $= 5.31 \times 10^{-26} \ kg$).

The ratio of de Broglie wavelengths associated with thermal neutrons at temperatures $127^{\circ} C$ and $352^{\circ} C$ is