Electrons are accelerated through a potential difference of $16 \ kV$. If the potential difference is increased to $64 \ kV$,then the de-Broglie wavelength associated with the electron will

If an electron and a photon propagate in the form of waves having the same wavelength,it implies that they have the same

$A$ particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is $1.813 \times 10^{-4}$. Calculate the particle's mass and identify the particle.

$A$ proton and an alpha particle are accelerated under the same potential difference. The ratio of de-Broglie wavelengths of the proton and the alpha particle is

$A$ 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:

What should be the order of arrangement of de-Broglie wavelength of an electron $(\lambda_{e})$,an $\alpha$-particle $(\lambda_{\alpha})$,and a proton $(\lambda_{p})$,given that all have the same kinetic energy?