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

If a proton is accelerated through a potential difference of $1000 \,V$, then its de-Broglie wavelength is (given, $m_p = 1.67 \times 10^{-27} \,kg$, $h = 6.63 \times 10^{-34} \,J-s$).

$A$ nucleus of mass $M$ at rest splits into two parts having masses $\frac{M'}{3}$ and $\frac{2M'}{3}$ (where $M' < M$). The ratio of the de Broglie wavelengths of the two parts will be:

$A$ 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).

An $\alpha$ particle and a proton are accelerated from rest by a potential difference of $200 \ V$. After this,their de Broglie wavelengths are $\lambda_{\alpha}$ and $\lambda_{p}$ respectively. The ratio $\frac{\lambda_{p}}{\lambda_{\alpha}}$ is

An electron of charge $e$ and mass $m$ moving with an initial velocity $v_0 \hat{i}$ is subjected to an electric field $E_0 \hat{j}$. The de-Broglie wavelength of the electron at a time $t$ is (Initial de-Broglie wavelength of the electron $= \lambda_0$)