An electron is accelerated from rest through a potential difference of $V$ volt. If the de Broglie wavelength of the electron is $1.227 \times 10^{-2} \, nm$,the potential difference is $......V$.

If the de Broglie wavelength of a particle is equal to the wavelength of a photon,then the energy of the photon is .....

The additional energy that should be given to an electron to reduce its de-Broglie wavelength from $1 \ nm$ to $0.5 \ nm$ is

$(a)$ For what kinetic energy of a neutron will the associated de Broglie wavelength be $1.40 \times 10^{-10} \; m ?$
$(b)$ Also find the de Broglie wavelength of a neutron,in thermal equilibrium with matter,having an average kinetic energy of $\frac{3}{2} k T$ at $300 \; K$.

The de Broglie wavelengths associated with a proton and an electron are in the ratio $2: 1$. Their stopping potentials are approximately in the ratio of

What should be the order of arrangement of de-Broglie wavelength of an electron $(\lambda_{e})$,an $\alpha$-particle $(\lambda_{\alpha})$,and a proton $(\lambda_{p})$,given that all have the same kinetic energy?