An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is ($c$ = velocity of light in air).

For an electron microscope,which of the following is false?

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

$A$ proton and an alpha particle are accelerated under the same potential difference. The ratio of de-Broglie wavelengths of the proton and the alpha particle is

If the de Broglie wavelength of a particle is equal to the wavelength of a photon,then the energy of the photon is .....

The potential energy of a particle of mass $m$ is given by $U(x) = \begin{cases} E_0; & 0 \le x \le 1 \\ 0; & x > 1 \end{cases}$. $\lambda_1$ and $\lambda_2$ are 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 $2 E_0$,the ratio $\frac{\lambda_1}{\lambda_2}$ will be: