The de-Broglie wavelength of an electron accelerated by a potential of $50\,V$ is approximately ............. $\mathring{A}$. $(|e| = 1.6 \times 10^{-19}\,C, m_e = 9.1 \times 10^{-31}\,kg, h = 6.6 \times 10^{-34}\,Js)$

An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are moving with the same speed. The ratio of their de-Broglie's wavelengths $\lambda_{e} / \lambda_{p}$ is

$A$ 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:

$A$ particle of mass $4M$ at rest disintegrates into two particles of mass $M$ and $3M$ respectively,having non-zero velocities. The ratio of the de-Broglie wavelength of the particle of mass $M$ to that of mass $3M$ will be:

The kinetic energy of an electron and a proton is $10^{-32} \ J$. Then the relation between their de-Broglie wavelengths is

If particles are moving with the same velocity,then the maximum de-Broglie wavelength will be for