The wavelength of a de-Broglie wave is $2 \mu m$. Calculate its momentum. (Given: $h = 6.63 \times 10^{-34} \ J \cdot s$)

The proton and $\alpha$-particle are accelerated through the same potential difference. Then the ratio of the de-Broglie wavelength of proton and $\alpha$-particle is (mass of $\alpha$-particle is $4$ times mass of proton, charge of $\alpha$-particle is $2$ times charge of proton).

The de-Broglie hypothesis treated electrons as

$A$ photon and an electron have equal energy $E$. The ratio $\lambda_{\text{photon}} / \lambda_{\text{electron}}$ is proportional to:

The de-Broglie wavelength of an electron moving with a velocity $v = c / 2$ ($c$ is the velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of the electron and the photon is:

The graph between the energy log $E$ of an electron and its de$-$Broglie wavelength log $\lambda$ will be