E = hc/λ equation proof Is there a proof for the equation of Photon Energy: $E = \frac{(h
\cdot c)}{\lambda}$? The teacher at school taught us that there's $E = h*f$, when I told him about $E = \frac{(h
\cdot c)}{\lambda}$, he said it was invalid.
So, is there any formula proof for this?
 A: It is not invalid; indeed since $f = c/\lambda$ one has for the energy of a photon $E = h c/\lambda$ by direct substitution. But your teacher might have wanted to be cautious because there is another equation like it, the de Broglie formula for momentum and wavelength, $p = h/\lambda.$ If you combine these very naively you get $E = p~c,$ a totally non-quantum result which is in fact true for light waves (as has been known since Maxwell discovered electromagnetic waves and conjectured that light was one of them, which was in the 1800s) but is not directly true for massive particles (for example a free nonrelativistic particle has $E = p^2/(2m)$ for its kinetic energy).
In fact in quantum mechanics we do see some reconciliation between the $E = h f$ and $p = h/\lambda$ viewpoints even for nonrelativistic particles but we have to be very clear on what we mean; usually $p = h/\lambda$ survives undisturbed as the way to calculate a wavelength of an electron, for example, but usually we also have two different states with two different energy-values, and we put a particle into a "superposition" of the two states: then generally the difference between their energy values $\Delta E$ usually corresponds to a frequency with which things are changing, so everything starts changing with frequency $\Delta E/h.$ Therefore the statement $E = h f$ is not 100% meaningful by itself for particles other than free photons. (You might have guessed this because usually the potential energy is only defined up to an additive constant and therefore only energy differences drive actual physics forward.)
