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

What will be the ratio of de-Broglie wavelengths of a proton and an $\alpha$-particle of the same energy?

The magnitude of the de-Broglie wavelength $(\lambda)$ of electron $(e)$,proton $(p)$,neutron $(n)$ and $\alpha-$ particle $(\alpha)$ all having the same energy of $1\,MeV$,in the increasing order will follow the sequence

Particle $A$ of mass $m_{A} = \frac{m}{2}$ moving along the $x$-axis with velocity $v_{0}$ collides elastically with another particle $B$ at rest having mass $m_{B} = \frac{m}{3}$. If both particles move along the $x$-axis after the collision,the change $\Delta \lambda$ in the de-Broglie wavelength of particle $A$,in terms of its de-Broglie wavelength $(\lambda_{0})$ before the collision is:

When the kinetic energy of an electron is increased,the wavelength of the associated wave will:

An $\alpha$-particle,a proton,and an electron have the same kinetic energy. Which one of the following is correct in the case of their De-Broglie wavelength?