# Planck's Constant, factor of 2

I'm just starting to look into the birth of Quantum mechanics, but I'm already a little confused by something.

I've seen 2 different equations involving the Planck's constant so far:

1. $hf = \mathrm{KE} - ϕ$

2. $h/p = λ \implies h/p = c/f \implies hf = cp \implies hf = 2\mathrm{KE}$

First of all, the way I understand these equations is they're talking about relationship between the energy of an electromagnetic wave the frequency of that wave(I'm not sure what it means by frequency yet, but when it says higher frequencies I'm just picturing the electron that produced the photon having moved faster in the atom, although IDK if thats correct)

Anyways, these equations for the kinetic energy seem to differ by a factor of 2...

What am I missing?

These are referring to two different things, I believe. In particular, this looks like it's discussing the photoelectric effect, where that light incident on a metallic surface ejects electrons from that surface by imparting energy to them. In that case, the first equation

$$hf = KE - \phi$$

is describing the relation between the energy of the photon emitted and the electron ejected. The $KE$ refers to the electron kinetic energy. Actually, this equation looks like it's for the inverse process (adsorption of an electron onto the metal surface followed by attendant release of a photon). The equation for ejection should be

$$KE = hf - \phi\ \mbox{(electron ejection)}$$

where in both cases $\phi$ is the work function of the material, that is, the energy required to desorb one electron from the surface and conversely the energy released when an electron is adsorbed. $hf$ is the photon energy of the incoming photon, part of which is expired to wrench the electron from the grip of the surface, while the remainder ends up as the kinetic energy in the outgoing electron. Note that if $hf < \phi$ then no electron ejection will occur (to first order).

The energy of a photon alone is just

$$E_\mathrm{photon} = hf$$

or

$$E_\mathrm{photon} = \hbar \omega$$

is preferred theoretically as angular velocity $\omega$ is considered more fundamental than frequency $f$.

I am not sure what the last equation equating to twice the kinetic energy of the electron (KE) is about. $E = pc$ is the relationship of photon energy to its momentum (the previous is to its frequency). $E = pc$ is actually from special relativity alone and applies to any massless object so also applies to a bulk light wave, it is not specifically quantum mechanical. The other energy-frequency relationships above are, however, as evidenced by the presence of $h$ in them, since photons are specifically part of quantum theory. (Rules of thumb: if an equation has $h$ or $\hbar$, quantum theory is involved somewhere. If it has $c$, relativistic theory is involved somewhere. If $G$ then gravitation or general relativity is involved somewhere.)

• Hey, thanks for your answer. I don't know much about quantum mechanics at all yet, but its just usually KE is pv/2 but now its pv or pc instead. I was just wondering about that. Aug 17, 2018 at 10:13

I just don't know where that facor comes out from. Are you using the virial theorem? If so, don't do it, because the origin of all this is a free particle, so there is no potential energy at all.

First of all, it was found that energy is $\hbar\omega$ (or $h\nu$ ). For a free particle, $E=E_k$.

Then, for a photon, you find that $E=pc$.

Therefore, you can show that

$E=pc = p \lambda \nu$

$$h \nu = p \lambda \nu$$ $$h=p\cdot \lambda$$ $$p=\frac{h}{\lambda} \equiv \frac{\hbar}{k}$$

So, finally, Broglie's hypotesis is: all particles fullfill these relations, not just photons.

$$E=h\nu = \hbar \omega$$ $$p=\frac{h}{\lambda} \equiv \frac{\hbar}{k}$$

• thanks for your answer. No, I'm not using any theorems, to be honest, I don't really know much about this yet. Its just normally the kinetic energy equation is E = pv/2, but in this case they're just saying its pv or pc (for light). Aug 17, 2018 at 10:12
• Ooh, okay, I see why your confusion now. A photon is described by relativistic equations. $E_k=½mv^2$ is actually an approximation of the real formula: $E^2=p^2c^2 + m_0^2 c^4$, but the photon has no mass. Aug 17, 2018 at 10:18
• Okay, okay. Hmm, I'm just gonna leave it at this for now and come back when I've learned more. I'm extremely confused, but its like I know so little right now I don't have the vocab to express my confusion :) However, one more thing: Do faster-oscillating electrons in the atom create higher energy photons? I think this would make sense, I understood it electromagnetism, was a way to look at electric attraction from a relativistic point of view, so that when electrons moved faster, the fields would contract more and therefore add up in the same space. Would it be the same with photons? Aug 17, 2018 at 10:30
• Well, I think you should learn about some special relativity basics before QM. At least until the famous formula $E=mc^2$ (where m is not the typical mass, this equation expands to the one I told you). As for your second issue, you should definitely ask a separate question. Comments must not extend so much. Aug 17, 2018 at 10:35