I'm going to have to give an answer that's very different to Jimmy360's. Apologies.

*How does the potential and kinetic energy of a photon relate?*

They don't. The photon is all kinetic energy.

*Do they mean the same thing?*

No. When you drop a brick, its gravitational potential energy is converted into kinetic energy. When you dissipate this kinetic energy as radiation, you're left with a mass deficit, see Wikipedia. From this you know that the potential energy was rest-mass energy. You also know that a photon doesn't have any rest mass, so you ought to know that potential energy doesn't apply.

*Also how does De broglie wavelength and potential relate?*

An electron has a de Broglie wavelength. When you drop the electron, its gravitational potential energy is converted into kinetic energy. When you dissipate this kinetic energy as radiation, the electron then has a mass deficit, and its de Broglie wavelength is increased.

People say that a descending photon is blueshifted, and that it gains energy. But I'm afraid it doesn't. Gravity is not a force in the Newtonian sense. If you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c². Conservation of energy applies. You're like the electron writ large. When you descend potential energy is converted into kinetic energy, which gets dissipated. So your total energy is reduced. So you measure the selfsame photon energy as increased. The frequency doesn't actually change, but gravitational time dilation means you and your clocks are going slower, so you *measure* the frequency as being increased.

You calculated the result of applying the momentum operator on the wave function. That's not the same as calculating the expectation value of the operator.

You'll have an easier time of it if you use the exponential version of the wavefunction $$e^{i(kx-\omega t)}$$

BTW, I think there's a typo in your momentum operator.

## Best Answer

To answer the question in your title - no, the kinetic energy of an electron is a function of its velocity, same as for any other particle. The

chargeof an electron is always $1.6\cdot 10^{-19}C$, and so the electron will pick up $1.6\cdot 10^{-19}J$ of energy for every Volt of accelerating potential.The key to answering the question (part d in the problem set) is the calculation of the wavelength of the electron.

The De Broglie wavelength is given by

$$\lambda = \frac{h}{p} = \frac{h}{mv}$$

Now the velocity is much less than the speed of light, so we don't need to consider relativistic effects. We know the rest mass of the electron, and we find

$$\lambda = \frac{6.6\cdot 10^{-34}}{9.1\cdot 10^{-31}\cdot 1.5\cdot 10^6}=0.5 nm$$

The other equation used ($p=\sqrt{2mK}$) appears to be using information not given in the question: namely that the kinetic energy of the electron is 5 eV, which is what you get when an electron is accelerated through 5 Volt. That has no visible bearing on the calculation of the wavelength, which can be derived directly from the velocity of the electron.

Incidentally, the momentum-from-energy calculation gives a completely different answer from the momentum-from-velocity equation, and indeed an electron with 5 eV of energy does not have a speed of 1.5E6 m/s... No wonder you were getting confused.