Momentum of a Photon: E/c=p Explained

[pokemon123]

Homework Statement

I have been asked to calculate the momentum of a photon in that has been ejected as a gamma ray after a nucleus was excited in terms of the Energy. I am confused as to whether or not I can use two different equations

Homework Equations

E=mc^2 or E=hf λ=h/p

The Attempt at a Solution

E=mc^2 E/c=mc
mv=p
c=v(of photon)
thus E/c=p

E=hf
f=v/λ
λ=hc/E
λ=h/p
h/p=hc/E
p=E/c

where I get confused is that the answer sheet only gives the second derivation as the answer and i have been told by someone that E=mc^2 is not the correct way to derive the momentum of a photon however I'm not sure why. Could someone please explain to me if both derivations are correct or if one is incorrect please explain why.

 

[Chandra Prayaga]

The mass of a photon is zero, so you cannot use E = mc². The second derivation is correct. The photon does carry both momentum and energy, and for these, both equations that you wrote in the second derivation are correct. E = hf and p = h/λ = E/c.

 

[Orodruin]

E=mc^2

To expand a bit on the previous reply, $E = mc^2$ is just an expression of the rest energy of an object. A photon has no rest frame and hence no rest energy. If you want to include the full relation between energy, momentum, and mass, you also need to include a term that depends on the momentum and the relation becomes
$$
E^2 = (pc)^2 + (mc^2)^2.
$$

mv=p

This is not a correct equation for a relativistic object. You need $p = m\gamma v$ and, as before, $m$ is zero for the photon but $\gamma$ is undefined as $\gamma \to \infty$ as $v\to c$. You get the correct answer in the end because generally you could introduce relativistic mass as a placeholder for total energy. The more general relationship between energy, momentum, and speed is $v = pc^2/E$, which is independent of the mass of the object and indeed gives $E = pc$ when $v = c$.