The accelerating voltage of an electron gun is $50,000 \ V$. What will be the de Broglie wavelength of the electron in $\mathring{A}$?

$A$ particle has a de Broglie wavelength $\lambda$. If its velocity and energy are both doubled,what will be the new de Broglie wavelength?

Which of the following statements is incorrect?

$A$ particle of charge $q$ and mass $m$ enters a region of a transverse electric field of $E_{0} \hat{j}$ with initial velocity $v_{0} \hat{i}$. The time taken for the change in the de-Broglie wavelength of the charge from the initial value of $\lambda_{0}$ to $\lambda_{0} / 3$ is proportional to

The de Broglie wavelength of an electron accelerated between two plates having a potential difference of $900 \ V$ is nearly. (in $nm$)

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5 \times 10^{-12} \ m$, the minimum electron energy required is close to .............. $keV$.