While going through de Broglie's relation in my textbook I was stuck by a derivation:
Here $$\ E=mc^2 $$ has been applied to photons. That confused me as I thought that mass of photon=0 but a photon does have energy. So I came across this post on SE. But this has confused me more. If the equation does not apply for photons then why have we used it here to derive the Broglie's relation?
Okay, so first, here's another good explanation for a photon's "mass":
...the word "mass" has been used in two different ways in physics. One was the way Einstein used it in $E=mc^2,$ where mass is really just the same thing as energy (E) but measured in different units. This is the same "m" that you multiply velocity by to find momentum (p), and thus is sometimes called the inertial mass. It's also the mass that provides the source of gravitational effects. Light has this "$m$" because it has energy.$^1$
This can be briefly summarized as follows:
There are two types of mass, the type you and I think about normally, like how much mass an apple has. The second type is what light has - inertial mass. This is the sort of mass Einstein used in $E=mc^2$, that really is just the same thing as energy, but in different units.
The equation does apply for photons. A way you can see this is true is the fact that light is affected by gravity - bent by the mass of stars/planets, and trapped by black holes. It isn't "against the rules" to use the definition of mass described above.
However, as Daniel Kerr says in the comments, de Broglie's derivation is for matter, so $E = mc^2$ definitely applies. The photon part isn't really related to what the book is talking about (at the very least, it makes it unclear).
I hope this helps!
$^1$This quote is from this website.
The argument in your book is heuristic, that is it presents a reasonable justification for the de Broglie wavelength but it is not a proof.
The correct expression for the de Broglie wavelength is:
$$ \lambda = \frac{h}{p} \tag{1} $$
For a massive particle in the non-relativistic limit the momentum is given by:
$$ p = mv $$
so we can write equation (1) as:
$$ \lambda = \frac{h}{mv} $$
For a photon we can use the relativistic equation linking energy and momentum:
$$ E^2 = p^2c^2 + m^2c^4 $$
And since for a photon $m=0$ this gives us $E=pc$ and therefore equation (1) becomes:
$$ \lambda = \frac{hc}{E} $$
Rearranging this gives the usual expression:
$$ E = \frac{hc}{\lambda} = h\nu $$
By comparison with a massive particle it is tempting to write the momentum of a photon as:
$$ p = mc $$
where $c$ is the photon velocity and $m$ is some hypothetical mass given by $E/c^2$. Indeed if we substitute $m=E/c^2$ in the above equation we get back the correct relativistic result:
$$ p = \frac{E}{c^2}c = \frac{E}{c} $$
But this is just numerology. The photon does not have a mass in any useful sense of the word.
The text "any material particle … by analogy with photons" shows that the author is talking about massless photons in contrast to mass particles.
That means that his following formulations are for the best imprecise. $ E = mc^2$ applies to mass which is rest energy, and it does not apply to other types of energy. Apparently, the author wanted to say:
"if the energy of the photon was rest energy, the following formula would apply"
and his objective was the derivation of the De-Broglie equation.
