The de-Broglie wavelength is proportional to

The voltage applied to an electron microscope to produce electrons of wavelength $0.50 \text{ Å}$ is (in $\text{ V}$)

$A$ particle has mass $m$. If its kinetic energy is increased to four times its initial value,what will be the change in its de Broglie wavelength?

If the de Broglie wavelengths of an electron and a proton are equal,then the kinetic energy of the electron is ........

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:

An electron of mass $m$ is moving in an electric field $\vec{E} = -2E_0\hat{i}$ $(E_0 = \text{constant} > 0)$, with an initial velocity $\vec{V} = v_0\hat{i}$ $(v_0 = \text{constant} > 0)$. If $\lambda_0 = \frac{h}{mv_0}$, its de Broglie wavelength at time $t$ is . . . . . . .