What is the additional energy that should be supplied to a moving electron to reduce its de Broglie wavelength from $1 \,nm$ to $0.5 \,nm$?

An electron (mass $m$) with initial velocity $\overrightarrow{v} = v_{0} \hat{i} + v_{0} \hat{j}$ is in an electric field $\overrightarrow{E} = -E_{0} \hat{k}$. If $\lambda_{0}$ is the initial de-Broglie wavelength of the electron,its de-Broglie wavelength at time $t$ is given by:

For the same objective,find the ratio of the least separation between two points to be distinguished by a microscope for light of $5000\,\mathring{A}$ and electrons accelerated through $100\,V$ used as the illuminating substance.

Kinetic energy of a proton is equal to energy $E$ of a photon. Let $\lambda_1$ be the de-Broglie wavelength of the proton and $\lambda_2$ be the wavelength of the photon. If $\frac{\lambda_1}{\lambda_2} \propto E^{n}$,then the value of $n$ is

If the kinetic energy of a particle is increased by $16$ times, the percentage change in the de Broglie wavelength of the particle is......... $\%$

The ratio of kinetic energies of a proton and an $\alpha$-particle is $16:1$. What is the ratio of their associated de Broglie wavelengths?