The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of $n$ is

$A$ particle of charge $q$,mass $m$,and energy $E$ has de-Broglie wavelength $\lambda$. For a particle of charge $2q$,mass $2m$,and energy $2E$,the de-Broglie wavelength is:

The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of $10^{-15} \;m$ or less. This structure was first probed in early $1970s$ using high energy electron beams produced by a linear accelerator at Stanford,$USA$. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron $= 0.511 \;MeV$.)

$A$ proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

The resolving power of an electron microscope is greater than the resolving power of an optical microscope because

$(a)$ 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.