The de-Broglie wavelength $(\lambda)$ of a particle is related to its kinetic energy $(E)$ as

What are called matter waves? Name the devices which employ the use of these matter waves.

If an electron and a proton have the same de-Broglie wavelength,then the kinetic energy of the electron is

$A$ light of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. The photoelectrons emitted from the surface have de-Broglie wavelength $\lambda_1$. Then the ratio $\lambda : \lambda_1^2$ is ($h =$ Planck's constant,$c =$ velocity of light,$m =$ mass of electron).

$A$ 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:

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: