B Atom's mass -- does it change with energy levels?

[jeremyfiennes]

TL;DR Summary

Does an atom in an excited state have a higher mass than when in its ground state?

Summary: Does an atom in an excited state have a higher mass than when in its ground state?

Summary: Does an atom in an excited state have a higher mass than when in its ground state?

Does an atom in an excited state have a higher mass than when in its ground state?

 

[DaveE]

Yes. m = E/c²

 

[jeremyfiennes]

So the photon which was absorbed and provided the additional excited state mass, also has mass.

 

[RPinPA]

So the photon which was absorbed and provided the additional excited state mass, also has mass.

No, the energy from the photon has been converted into mass.

 

[weirdoguy]

Yes. m = E/c2

Assuming that we are working in a reference frame in which atom is at rest. There is no frame in which photon is at rest, so @jeremyfiennes you can't use this equation to describe a photon. General formula looks like this: $m=\frac{1}{c^2}\sqrt{E^2-c^2\vec{p}^2}$. For a single photon $E=c|\vec{p}|$, so $m=0$.

 

[jeremyfiennes]

So E=mc^2 doesn't apply to photons, which have energy but no mass?

 

[weirdoguy]

$mc^2$ is called rest energy and should be written as $E_0$ with a subscript. And only in a frame in which object is at rest we have $E=E_0$. Photons are never at rest so we can't talk about their rest energy. General formula involves momentum, as I stated in my last post.

 

[jeremyfiennes]

Electrons orbiting at 99.9% of the speed of light have energy and corresponding E=mc^2 mass. So why don't photons traveling at 100% of the speed of light. I don't get the distinction.

 

[weirdoguy]

Electrons orbiting at 99.9% of the speed of light have energy and corresponding E=mc^2 mass.

No. Putting aside the fact, that in QM you can't talk about the speed of electron orbiting a nucleus, in a reference frame in which electron has speed $0.999c$ you can't use $E=mc^2$. You have to use general formula $m=\frac{1}{c^2}\sqrt{E^2-c^2\vec{p}^2}$, because in this reference frame momentum $\vec{p}$ is non-zero. Read carefully what I wrote:

And only in a frame in which object is at rest we have $E=E_0$.

 

[jeremyfiennes]

Thanks. I will have to think. In the meantime: why can't one talk of the speed of an electron?

 

[Nugatory]

Thanks. I will have to think. In the meantime: why can't one talk of the speed of an electron?

You can, and I’m not seeing where anyone said you can’t.

What you cannot do is use the classical formulas $p=mv$ (momentum) and $E=mv^2/2$ (kinetic energy) when $v$ is not small compared with $c$, nor the relativistic $E=mc^2$ when $v$ is non-zero.

 

[RPinPA]

Thanks. I will have to think. In the meantime: why can't one talk of the speed of an electron?

You can talk about the speed of an electron. You can't talk about the speed of an electron orbiting a nucleus, because the electron isn't orbiting the nucleus like a little planet. That's the Bohr model of the atom and it is known not to be a valid description.

Electrons orbiting at 99.9% of the speed of light have energy and corresponding E=mc^2 mass.

In the expression $mc^2$, $m$ for most physicists these days is what used to be called the "rest mass". So the expression $E = mc^2$ is the amount of energy the electron has when it is at rest.

When it's moving, it has kinetic energy in addition to the rest energy.

But that energy is bound up in a sense in the form of the mass. If the electron meets a positron, both particles will annihilate and then you'll have the energy available. But the mass will be gone.

So why don't photons traveling at 100% of the speed of light. I don't get the distinction.

Photons are missing the "rest energy" because photons are never at rest.

 

[Dale]

So the photon which was absorbed and provided the additional excited state mass, also has mass.

No, mass isn’t additive. The mass of a combined object is equal or greater than the masses of its constituents.

 

[weirdoguy]

why can't one talk of the speed of an electron?

You really should pay attention to every word in a sentence... In the beggining I was talking about electron in a quantum mechanical atom, not free electron. In an atom things are not that easy.