E = Hc/λ Equation Proof - Physics Stack Exchange

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Learn more about Teams E = hc/λ equation proof Ask Question Asked 7 years, 11 months ago Modified 7 years, 11 months ago Viewed 13k times 2 $\begingroup$

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?

• energy
• photons

Share Cite Improve this question Follow edited Jan 12, 2017 at 22:18 auden 7,08744 gold badges3333 silver badges6060 bronze badges asked Jan 12, 2017 at 22:08 Coto TheArcherCoto TheArcher 17511 gold badge22 silver badges55 bronze badges $\endgroup$ 2

• 2 $\begingroup$ take what you know and use $\lambda=c/f$. $\endgroup$ – AccidentalFourierTransform Commented Jan 12, 2017 at 22:10
• $\begingroup$ @AccidentalFourierTransform How would I use that to solve it as E? $\endgroup$ – Coto TheArcher Commented Jan 12, 2017 at 22:11

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Sorted by: Reset to default Highest score (default) Date modified (newest first) Date created (oldest first) 2 $\begingroup$

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.)

Share Cite Improve this answer Follow answered Jan 12, 2017 at 22:47 CR DrostCR Drost 39k33 gold badges4343 silver badges116116 bronze badges $\endgroup$ 2

• 2 $\begingroup$ $E=hc/\lambda$ is true, but only if we point out that $\lambda$ denotes the vacuum wavelength of the photon, not the wavelength in whatever material the photon is travelling in. I suspect this is the distinction OP's instructor was trying to be careful about. $\endgroup$ – The Photon Commented Jan 13, 2017 at 16:42
• $\begingroup$ Turns out I gave him (h*(c^2))/λ instead of (h*c)/λ. I showed him today and he said "Yes, this is correct, not the squared one you'd showed me yesterday". $\endgroup$ – Coto TheArcher Commented Jan 13, 2017 at 18:24

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