The ratio of the de-Broglie wavelengths for the electron and proton moving with the same velocity is ($m_p$ = mass of proton,$m_e$ = mass of electron).

If $\lambda_0$ is the de-Broglie wavelength for a proton accelerated through a potential difference of $100 \ V$,the de-Broglie wavelength for an $\alpha$-particle accelerated through the same potential difference is

$A$ 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

The de Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is ...... .

$A$ particle is moving along the $x-$axis back and forth in a box of length $L$. Assuming the de-Broglie hypothesis is applicable for the particle,and it is moving with a constant speed in the box making perfectly elastic collisions with the walls. The possible value of the momentum of the particle is (Note: $h$ is Planck's constant,$n$ is the principal quantum number).

The ratio of momentum of an electron and an alpha particle which are accelerated from rest by a potential difference of $100\,V$ is