If the momentum of an electron is changed by $\Delta p,$ then the de-Broglie wavelength associated with it changes by $0.50\%.$ The initial momentum of the electron will be:

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is:

$A$ particle has mass $m$. If its kinetic energy is increased to four times its initial value,what will be the change in its de Broglie wavelength?

Two large parallel plates are connected to a $100 \, V$ power supply. These plates have a fine hole at the centre. An electron having energy $200 \, eV$ is directed such that it passes through the holes. When it emerges, its de-Broglie wavelength is ............. $\mathring{A}$

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

$A$ proton,a neutron,an electron,and an $\alpha$-particle have the same energy. If $\lambda_{p}, \lambda_{n}, \lambda_{e},$ and $\lambda_{\alpha}$ are the de Broglie wavelengths of the proton,neutron,electron,and $\alpha$-particle respectively,then choose the correct relation from the following: