$A$ particle is moving $5$ times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is $1.878 \times 10^{-4}$. The mass of the particle is close to

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:

$A$ 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:

$A$ particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is head-on and elastic. The ratio of the de-Broglie wavelengths $\lambda_A$ and $\lambda_B$ after the collision is

If the linear momentum of a proton is changed by $p_0$,then the de Broglie wavelength associated with the proton changes by $0.25 \%$. Then the initial linear momentum of the proton is

The energy that should be added to an electron to reduce its de-Broglie wavelength from $1 \,nm$ to $0.5 \,nm$ is