The kinetic energy of an electron with a de-Broglie wavelength of $0.3 \, nm$ is ............. $eV$.

State the de-Broglie hypothesis.

The de-Broglie wavelength of an electron having $80 \ eV$ of energy is nearly .............. $\mathring{A}$ ($1 \ eV = 1.6 \times 10^{-19} \ J$,Mass of electron $= 9 \times 10^{-31} \ kg$,Planck's constant $= 6.6 \times 10^{-34} \ J \cdot s$).

Two particles move at right angles to each other. Their de Broglie wavelengths are $\lambda_1$ and $\lambda_2$ respectively. The particles undergo a perfectly inelastic collision. The de Broglie wavelength $\lambda$ of the final particle is given by

An alpha particle moves along a circular path of radius $0.5 \ mm$ in a magnetic field of $2 \times 10^{-2} \ T$. The de Broglie wavelength associated with the alpha particle is nearly (Planck's constant $= 6.63 \times 10^{-34} \ J \ s$)

An electron (of mass $m$) and a photon have the same energy $E$ in the range of a few $eV$. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is ($c =$ speed of light in vacuum).