The de-Broglie wavelength of a particle having kinetic energy $E$ is $\lambda$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to $75 \%$ of the initial value?

The wavelength of light from the spectral emission line of sodium is $589 \; nm$. Find the kinetic energy at which
$(a)$ an electron,and
$(b)$ a neutron,would have the same de Broglie wavelength.

The proton and $\alpha$-particle are accelerated through the same potential difference. Then the ratio of the de-Broglie wavelength of proton and $\alpha$-particle is (mass of $\alpha$-particle is $4$ times mass of proton, charge of $\alpha$-particle is $2$ times charge of proton).

The magnitude of the de-Broglie wavelength $(\lambda)$ of electron $(e)$,proton $(p)$,neutron $(n)$ and $\alpha-$ particle $(\alpha)$ all having the same energy of $1\,MeV$,in the increasing order will follow the sequence

The de-Broglie wavelength of a proton (charge $= 1.6 \times 10^{-19} \ C$,mass $= 1.67 \times 10^{-27} \ kg$) accelerated through a potential difference of $1 \ kV$ is:

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