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:

If the potential difference used to accelerate electrons is doubled,by what factor does the de Broglie wavelength $(\lambda)$ associated with the electrons change?

$A$ proton and an $\alpha$-particle are accelerated from rest to the same kinetic energy. The ratio of their de Broglie wavelengths $\lambda_{p} : \lambda_{\alpha}$ is:

An electron is moving through a field. It is moving $(i)$ opposite to an electric field and $(ii)$ perpendicular to a magnetic field as shown. For each situation, determine the de-Broglie wavelength of the electron:

If an electron and a photon propagate in the form of waves having the same wavelength,it implies that they have the same

We wish to see inside an atom. Assuming the atom to have a diameter of $100 \ pm$, this means that one must be able to resolve a width of say $10 \ pm$. If an electron microscope is used, the minimum electron energy required is about....... $KeV$.