$A$ particle of charge $q$,mass $m$,and energy $E$ has de-Broglie wavelength $\lambda$. For a particle of charge $2q$,mass $2m$,and energy $2E$,the de-Broglie wavelength is:

An electron beam of energy $10 \, KeV$ is incident on a metallic foil. If the interatomic distance is $0.2454 \, \mathring{A}$,then find the angle of diffraction in degrees. $(\sqrt{150} = 12.27)$

The energy that should be added to an electron to reduce its de Broglie wavelength from $1 \, nm$ to $0.5 \, nm$ is

The speed of a proton is $c / 20$. What will be the de Broglie wavelength associated with it?

An electron of mass $m$ with an initial velocity $\vec{V} = V_0 \hat{i} \,(V_0 > 0)$ enters an electric field $\vec{E} = -E_0 \hat{i} \,(E_0 = \text{constant} > 0)$ at $t = 0$. If $\lambda_0$ is its de-Broglie wavelength initially,then its de-Broglie wavelength at time $t$ is:

The de-Broglie wavelength $(\lambda)$ of a particle