The wavelength associated with an electron accelerated through a potential difference of $100 \ V$ is nearly .............. $ \mathring A $

$A$ proton,a neutron,an electron,and an $\alpha$-particle have the same energy. If $\lambda_{p}, \lambda_{n}, \lambda_{e},$ and $\lambda_{\alpha}$ are the de Broglie wavelengths of the proton,neutron,electron,and $\alpha$-particle respectively,then choose the correct relation from the following:

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

What is the de-Broglie wavelength of the $\alpha$-particle accelerated through a potential difference $V$?

$A$ particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is head-on and elastic. The ratio of the de-Broglie wavelengths $\lambda_A$ and $\lambda_B$ after the collision is

$A$ proton and a photon both have the same energy of $E = 100 \,keV$. If the de Broglie wavelength of the proton is $\lambda_1$ and that of the photon is $\lambda_2$,then $\lambda_1/\lambda_2$ is proportional to: