If the kinetic energy of a moving particle is $E$,then the De Broglie wavelength is

$A$ proton and an $\alpha$-particle are accelerated through a potential difference of $100\, V$. The ratio of the wavelength associated with the proton to that associated with an $\alpha$-particle is

What voltage should be applied to an electron microscope to produce electrons with a wavelength of $\lambda = 1.0 \ \mathring{A}$ (in $V$)?

An electron of mass $m$ with an initial velocity $\overrightarrow{v}=v_0 \hat{i}$ $(v_0>0)$ enters an electric field $\overrightarrow{E}=-E_0 \hat{k}$. If the initial de Broglie wavelength is $\lambda_0$,the value after time $t$ would be $:-$

When the kinetic energy of an electron is increased,the wavelength of the associated wave will:

Find the de Broglie wavelength of an electron in a metal at $27^{\circ}C$.