Question

In class we deduced the following:
E=mc^2 and for photons we know: E=hf
therefore m=hf/c^2 and if we plug that into p=mv we get p=hf/c^2 * c = hf/c or p=h/lambda. That came all from the fact that photons have no rest mass.
Later we used p=h/lambda for electrons. Isn´t that wrong because we deduced that from a particle with no rest mass but an electron has a rest mass and therefore this formula should not work?
I think we also used it to calculate the wavelength of a baseball with the impulse p=1 to find out that it is way to small.

Answer 1

De-Broglie couldn't be wrong(or can't be proved wrong so easily!). So there must be a fault in your derivation.
Personally, I think E=mc^2 shows how much energy we get when a mass m is converted to energy. It is not valid for a photon(because it has no mass). So E=mc^2 and E=hf cannot be used at the same place. So far, clear?
If you want to know how De-Broglie actually got that p=h/lamba, you should google and search for his original paper.

Answer 2

Photons have no rest mass but they have a mass or why can´t I say hf=mc^2? The move with the speed of light and stuff :D

Answer 3

They do. But to apply E=mc^2, you don't need your particle to move at the speed of light. It happens in nuclear fission and fusion, without the speed of light. But to apply E=hv, you need electromagnetic radiation.
Though Eienstein developed both these relations, he developed the second one while studying photoelectric effect, and the first one while thinking of mass-energy equivalence.
Take a look at the De-broglie equation here(scroll down a bit):
http://en.wikipedia.org/wiki/De_Broglie_hypothesis
You see that the second form of the equation asks for rest mass, which a photon doesn't have.

Answer 4

Isn´t a photon electromagnetic radiation?

Answer 5

Yes. It is the simplest unit of radiation. You can use E=hv for photons.

Answer 6

OKAY OKAY OKAY>>>>

what does E=mc^2 represent first of all? Energy 'course. But what kinda energy? Energy derived from where?

Answer 7

As far as I know E=mc^2 is the sum of the energy a body has due to its rest mass and its kinetic energy -> mc^2=E(kin) + m0c^2. (m0 = rest mass)
The photons have no m0c^2 term because their rest mass is zero so there e=mc^2=e(kin).

Answer 8

Lol no Tom. If a 1 kg weight near you had energy 1*(3*10^*)^2 Joules, all the world's energy crisis could've been solved!
E=mc^2 denotes the amount of energy that would be released when a mass m gets completely converted into energy.

Answer 9

I'm not sure about the premise of your question Tom
E=hf
does not come from the fact that the particle has no rest mass
the waves are quantum mechanical, and exist for electrons as well, as well as for the baseball (only they are much too small to measure) as your teacher stated.
I'm not sure why you think that formula is invalid for electrons?

Answer 10

:) @Mani_Jha I know that but is also a way to calculate the whole energy of something and another way would be using E=hf.

@TuringTest
I think the formula is invalid because for photons the whole energy is given by hf or mc^2 so we deduced that m=hf/c^2 for photons and with that we concluded that the impulse of a photon is given through the formula p=h/lambda.
Then we used the same formula for electrons but for electrons...
One of those cases where on one day it makes no sense and then the day after it seems clear. :)

Thx @everyone for the discussion
I guess that helped a lot because I made me think about it in more detail.