The voltage applied to an electron microscope to produce electrons of wavelength $0.50 \text{ Å}$ is (in $\text{ V}$)

An electron is accelerated from rest through a potential difference of $V$ volt. If the de Broglie wavelength of the electron is $1.227 \times 10^{-2} \, nm$,the potential difference is $......V$.

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 de Broglie wavelength of an electron moving with kinetic energy of $144 \;eV$ is nearly

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:

If $\lambda$ and $K$ are the de Broglie wavelength and kinetic energy,respectively,of a particle with constant mass,the correct graphical representation for the particle will be: