$A$ particle of charge $q$,mass $m$,and energy $E$ has de-Broglie wavelength $\lambda$. For a particle of charge $2q$,mass $2m$,and energy $2E$,the de-Broglie wavelength is:

The ratio of de Broglie wavelengths of a proton and an $\alpha$-particle having the same energy is ............

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

$A$ proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

The graph which shows the variation of the de Broglie wavelength $(\lambda)$ of a particle and its associated momentum $(p)$ is

The ratio of the de-Broglie wavelength of an $\alpha$-particle to that of a proton,when both are subjected to the same magnetic field such that the radii of their paths are equal,assuming the magnetic field induction vector $\vec{B}$ is perpendicular to the velocity vectors of the $\alpha$-particle and the proton,is: