The wavelength of a charged particle of mass $8.0 \times 10^{-31} \ kg$, charge $1.6 \times 10^{-19} \ C$ and kinetic energy $3 \ keV$ will be (Planck constant, $h = 6.4 \times 10^{-34} \ Js$) (in $\text{Å}$)

The de Broglie wavelength of an electron in a metal at $27^{\circ}C$ is compared with the given mean distance between two electrons in the metal,which is $2 \times 10^{-10} \ m$. The ratio of the mean distance to the de Broglie wavelength is approximately:

What is the de-Broglie wavelength associated with an electron moving with a speed of $6.4 \times 10^{6} \ m/s$ (in $nm$)? $(m_{e} = 9.11 \times 10^{-31} \ kg, h = 6.63 \times 10^{-34} \ J \cdot s)$

An electron microscope works on the principle of......

$A$ charged particle $q_0$ of mass $m_0$ is projected along the $y-$axis at $t = 0$ from the origin with a velocity $v_0$. If a uniform electric field $E_0$ also exists along the $x-$axis,then the time at which the de Broglie wavelength of the particle becomes half of its initial value is:

If the kinetic energy of a free electron is doubled,by what factor does its de Broglie wavelength change?