If a photon,an electron,and a uranium nucleus have the same de Broglie wavelength,which one will have the highest energy?

The de Broglie wavelength of a charged particle accelerated through a potential difference $V$ is $\lambda$. If the potential difference is increased by $21 \%$,the de Broglie wavelength of the charged particle is

An electron microscope uses electrons accelerated by a voltage of $50\; kV$. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture,etc.) are taken to be roughly the same,how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

We wish to see inside an atom. Assuming the atom to have a diameter of $100 \ pm$, this means that one must be able to resolve a width of say $10 \ pm$. If an electron microscope is used, the minimum electron energy required is about....... $KeV$.

An electron,a doubly ionized helium ion $(He^{++})$ and a proton have the same kinetic energy. The relation between their respective de-Broglie wavelengths $\lambda_{e}, \lambda_{He^{++}}$ and $\lambda_{P}$ is:

Proton $(P)$ and electron $(e)$ will have the same de Broglie wavelength when the ratio of their momentum is (assume,$m_{p} = 1849 \, m_{e}$)