The ratio of momenta of an electron and an $\alpha$-particle which are accelerated from rest by a potential difference of $100 \ V$ is

If a proton and an $\alpha$-particle have the same kinetic energy,what is the ratio of their de-Broglie wavelengths?

$A$ particle is travelling $4$ times as fast as an electron. Assuming the ratio of the de-Broglie wavelength of the particle to that of the electron is $2:1$,the mass of the particle is:

The speed of electrons in a scanning electron microscope is $1 \times 10^{7} \, m/s$. If protons having the same speed are used instead of electrons,then the resolving power of the scanning proton microscope will be changed by a factor of:

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:

The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is $E$. If $\lambda_1$ and $\lambda_2$ are the de Broglie wavelengths of the proton and the photon respectively,then find the ratio $\lambda_1 / \lambda_2$.