The ratio of the momentum of an electron to that of an alpha particle, both accelerated from rest through the same potential difference $V$, is:

The $log-log$ graph between the energy $E$ of an electron and its de-Broglie wavelength $\lambda$ will be

An $\alpha$-particle and a proton are accelerated from rest by the same potential. The ratio of their de-Broglie wavelengths is . . . . . .

An electron of mass $m$ with an initial velocity $\overrightarrow{v}=v_0 \hat{i}$ $(v_0>0)$ enters an electric field $\overrightarrow{E}=-E_0 \hat{k}$. If the initial de Broglie wavelength is $\lambda_0$,the value after time $t$ would be $:-$

The resolving power of an optical instrument depends on the wavelength of the light used. An electron microscope uses a beam of electrons. Its resolving power can be increased if we:

The potential energy of a particle of mass $m$ is given by $U(x) = \begin{cases} E_0, & 0 \le x \le 1 \\ 0, & x > 1 \end{cases}$. Let $\lambda_1$ and $\lambda_2$ be the de-Broglie wavelengths of the particle when $0 \le x \le 1$ and $x > 1$ respectively. If the total energy of the particle is $2E_0$,find $(\lambda_1/\lambda_2)^2$.