# Momentum of an electron acting as a wave [closed]

Was working on a problem with electrons acting as waves in diffraction. Part of the question asked me to calculate the momentum of the electron. Since I was dealing with waves I used the following equation:

$h=pλ \implies p = h/λ$

Since $λ = v/f$ we can substitute that in, resulting in $p = hf/v$.

Substituting in the de Broglie $h = E/f$ into the above equation we get $p = E/v$. Since we're talking about electrons the only energy that the electron has is kinetic so we can substitude $E = 0.5mv^2$ into the equation giving us $p = 0.5mv^2/v = 0.5mv$.

I've repeat that, $p = 0.5mv$. Any 4th-grade physicist knows that momentum is $mv$ so on one hand, I have mv and on the other I have a derivation saying the momentum is $0.5mv$. Is there a mistake in my derivation I'm not seeing?

P.S: I noticed something a bit later. $p = E/v \implies E = pv = mv^2$. See any similarities between this and another infamous equation in the realm of relativity?

• $E=hf$ applies to photons. It is not valid for electrons. Commented Jan 13, 2017 at 14:29

## 2 Answers

From Einstein's famous equation we have,

$E=mc^2$

From classical mechanics we have,

$E = \frac{mv^2}{2}$

Equate both of them (both are E, right?) and you'll get

$mc^2 = \frac{mv^2}{2}$

$c^2 = \frac{v^2}{2}$

All objects in the universe are moving at $c\sqrt{2}$

Yes, they are moving FASTER than light. Oh dear! Physics does not work!

Do you notice the mistake?

You cannot equate arbitrary equations with one another even if the quantity has the same sign and is of the same type.

$E = h\nu$ is an equation which describes the energy carried by an electromagnetic wave of frequency $\nu$

Are electrons photons?

No. How can you equate the kinetic energy of an electron with the energy associated with a photon of a particular frequency? Does that even make sense?

• Wow, I was this close to downvoting you. Good thing I kept reading :) Commented Jan 13, 2017 at 19:40
• According classical mechanics we have other types of energy also . Commented Feb 24, 2017 at 12:46

Since I was dealing with waves I used the following equation: $$h=pλ \implies p = h/λ$$

This is fine.

Since $λ = v/f$ ...

True for all waves, yes.

... we can substitute that in, resulting in $p = hf/v$.

This is okay, but we don't usually find people talking about the "frequency" of an electron matter wave.

Substituting in the de Broglie $h = E/f$ ...

Ah, here's your problem. The de Broglie equation has nothing to do with energy or frequency --- the de Broglie relation is $p = h/\lambda$. The relation $E=hf$ is a result for massless particles, see below.

... Since we're talking about electrons the only energy that the electron has is kinetic ...

This is a reasonable statement for a free particle.

... so we can substitude $E = 0.5mv^2$

Nope, that's a nonrelativistic approximation muddled by a confusing notation. The relativistically correct statement is $$E^2 = (pc)^2 + (mc^2)^2 \tag 1$$ where $E$ is the total energy, including the rest energy, and the kinetic energy is $K = E-mc^2$. For massless particles, the Einstein equation (1), the de Broglie equation $p=h/\lambda$, and the wave equation result $\lambda f = v$ combine to give the result $E=hf$. However for massive particles you find instead \begin{align} E^2 &= \left(\frac{hfc}{v}\right)^2 + (mc^2)^2 \\ K &= -mc^2 + \sqrt{\left({hfc}/{v}\right)^2 + (mc^2)^2} \end{align} which is much less helpful. Hence people don't talk much about the frequncies of matter waves.