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

If the velocity of a moving particle is reduced to half, what will be the percentage change in its de Broglie wavelength?

Photons of wavelength $\lambda$ are incident on the cathode of a photocell. Electrons are emitted from the cathode surface. The de-Broglie wavelength of the emitted electrons is (work function is negligible).
($c =$ velocity of light,$h =$ Planck's constant,$m =$ mass of electron)

If the de Broglie wavelength of a neutron at a temperature of $77^{\circ} C$ is $\lambda$,then the de Broglie wavelength of the neutron at a temperature of $1127^{\circ} C$ is

The equation $\lambda = \frac{1.227}{x} \text{ nm}$ can be used to find the de-Broglie wavelength of an electron. In this equation,$x$ stands for:

Compute the typical de Broglie wavelength of an electron in a metal at $27\,^{\circ} C$ and compare it with the mean separation between two electrons in a metal which is given to be about $2 \times 10^{-10} \; m$.