The de-Broglie wavelength of an electron (mass $= 1 \times 10^{-30} \ kg$, charge $= 1.6 \times 10^{-19} \ C$) with a kinetic energy of $200 \ eV$ is (Planck's constant $= 6.6 \times 10^{-34} \ J \cdot s$):

The wavelength $\lambda_e$ of an electron and $\lambda_p$ of a photon of same energy $E$ are related by

The ratio of the de-Broglie wavelengths for the electron and proton moving with the same velocity is ($m_p$ = mass of proton,$m_e$ = mass of electron).

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

$A$ material particle with a rest mass $m_0$ is moving with the velocity of light $C$. Then,the wavelength of the de$-$Broglie wave associated with it is

An electron (mass $m$) with an initial velocity $\overrightarrow{v} = v_{0} \hat{i} \left(v_{0} > 0\right)$ is moving in an electric field $\overrightarrow{E} = -E_{0} \hat{i} \left(E_{0} > 0\right)$ where $E_{0}$ is constant. If at $t = 0$ the de Broglie wavelength is $\lambda_{0} = \frac{h}{mv_{0}}$,then its de Broglie wavelength after time $t$ is given by: