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

The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of $n$ is

An $\alpha$-particle and a deuteron are moving with velocities $v$ and $2v$ respectively. What will be the ratio of their de Broglie wavelengths?

What does the de Broglie equation for an electron represent?

$A$ beam of electrons of energy $E$ scatters from a target having atomic spacing of $1 \, Å$. The first maximum intensity occurs at $\theta = 60^{\circ}$. Then $E$ (in $eV$) is: (Planck constant $h = 6.64 \times 10^{-34} \, Js$, $1 \, eV = 1.6 \times 10^{-19} \, J$, electron mass $m = 9.1 \times 10^{-31} \, kg$)

$A$ particle has mass $m$. If its kinetic energy is increased to four times its initial value,what will be the change in its de Broglie wavelength?