An $\alpha$-particle and a proton are accelerated from rest by a potential difference of $100 \ V$. After this,their de Broglie wavelengths are $\lambda_\alpha$ and $\lambda_{p}$ respectively. The ratio $\frac{\lambda_{p}}{\lambda_\alpha}$,to the nearest integer,is

Electrons used in an electron microscope are accelerated by a voltage of $25 \; kV$. If the voltage is increased to $100 \; kV$,then the de-Broglie wavelength associated with the electrons would

$A$ bomb projected from the ground at an angle $\theta$ $\left( \theta \neq 90^\circ \right)$ explodes into two fragments of equal mass at the topmost point of its trajectory. If one of the fragments returns to the point of projection,then the ratio of the de Broglie wavelength of the second fragment just after the explosion to that of the bomb just before the explosion is:

Electrons used in an electron microscope are accelerated by a voltage of $25 \ kV$. If the voltage is increased to $100 \ kV$,then the de-Broglie wavelength associated with the electrons would

$A$ photon and an electron have equal energy $E$. The ratio $\lambda_{\text{photon}} / \lambda_{\text{electron}}$ is proportional to:

The de-Broglie wavelength is proportional to