The de-Broglie wavelength of a proton (charge $= 1.6 \times 10^{-19} \ C$,mass $= 1.67 \times 10^{-27} \ kg$) accelerated through a potential difference of $1 \ kV$ is:

$A$ long pipe is filled with small granules of nickel. The interatomic spacing for nickel is $d$. $A$ beam of electrons having different energies is incident on the left flat face of the pipe. Mass of electron is $m$. The beam of electrons coming out of the right flat face has energy:

The de-Broglie wavelength associated with a particle of mass $m$ moving with velocity $v$ is:

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

What is the de Broglie wavelength of an electron accelerated through $80 \ V$ in $\mathring A$?

An electron with mass $m$ and an initial velocity $(t=0)$ $\vec{v} = v_0 \hat{i}$ $(v_0 > 0)$ enters a magnetic field $\vec{B} = B_0 \hat{j}$. If the initial de Broglie wavelength at $t=0$ is $\lambda_0$,then its value after time $t$ would be: