If we consider an electron and a photon of the same de-Broglie wavelength,then they will have the same:

The de Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is ...... .

Assuming the nitrogen molecule is moving with $r.m.s.$ velocity at $400 \ K$, the de$-$Broglie wavelength of the nitrogen molecule is close to $...... \ \mathring{A}$. (Given: nitrogen molecule mass: $4.64 \times 10^{-26} \ kg$, Boltzmann constant: $1.38 \times 10^{-23} \ J/K$, Planck constant: $6.63 \times 10^{-34} \ J \cdot s$)

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

If the de-Broglie wavelength at a temperature of $27^oC$ is $\lambda$,what will be the de-Broglie wavelength at $927^oC$?

Write a note on the electron microscope.