If the de Broglie wavelength of a dust particle of mass $1.0 \times 10^{-9} \,kg$ is $3 \times 10^{-25} \,m$, then the speed of the particle is . . . . . . . $\left(h=6.625 \times 10^{-34} \,J \,s\right)$

$A$ proton and an electron have the same de Broglie wavelength. If $K_p$ and $K_e$ are the kinetic energies of the proton and electron respectively,then choose the correct relation:

When the kinetic energy of an electron is increased,the wavelength of the associated wave will

Compute the typical de Broglie wavelength of an electron in a metal at $27\,^{\circ} C$ and compare it with the mean separation between two electrons in a metal which is given to be about $2 \times 10^{-10} \; m$.

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....... $KeV$.

The ratio of the accelerating potentials required to accelerate $(i)$ an $\alpha$-particle and (ii) a proton to have the same de Broglie wavelength associated with them is (mass of $\alpha$-particle = $6.4 \times 10^{-27} \text{ kg}$, mass of proton = $1.6 \times 10^{-27} \text{ kg}$)