$A$ proton and an electron are accelerated through the same potential difference. The ratio $\frac{\lambda_e}{\lambda_p}$ will be:

When an electron is accelerated through a potential $V$,the de-Broglie wavelength associated with it is $4 \lambda$. When the accelerating potential is increased to $4V$,its wavelength will be

The de Broglie wavelength of a proton is twice the de Broglie wavelength of an alpha particle. The ratio of the kinetic energies of the proton and the alpha particle is

The de Broglie wavelengths for an electron and a photon are $\lambda_{e}$ and $\lambda_{p}$ respectively. For the same kinetic energy $K$ of the electron and the photon,which of the following presents the correct relation between the de Broglie wavelengths of the two?

$A$ 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:

The de Broglie wavelength $\lambda$ associated with a proton increases by $25 \%$,if its momentum is decreased by $p_0$. The initial momentum was