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The de Broglie wavelength equation is given by: $$\lambda = \frac{h}{mv}$$

Even though this is used to find the wavelength of the matter (matter waves) , I was curious to find the mass of photon through this since it has been experimentally found it has about $10^{-54}$ kg rest mass and whether I could reproduce it with the equation.

So $$m=\frac{h}{\lambda v}$$ plugging in $h=6.626\cdot 10^{-34} m^{2}kgs^{-1}$
considering light to be of red wavelength $\lambda = 700 \cdot 10^{-9} m $
and $v=c=3\cdot 10^{8} ms^{-1}$ I get a value about $3\cdot 10^{-36} $kg which is not even close to the experimental value , where did I go wrong here ?

Frederic Thomas
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Naveen V
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    Photons are massless particles. Where did you find a mass of $10^{-54}$ kg? – Aiden Mar 15 '23 at 07:31
  • @Aiden https://www.sciencedirect.com/science/article/pii/S2211379719330943#:~:text=According%20to%20electromagnetic%20theory%2C%20the,25%20MeV%20c%20%2D%202%20)%20. imma just add it to the main question too – Naveen V Mar 15 '23 at 07:31
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    The direct paper you linked is a horrendous mess in broken English, and the experiments it cites only give an upper bound on a hypothetic rest mass of the photon. By all experimental means and modern theory, the rest mass of the photon is zero. – John Doe Mar 15 '23 at 07:44
  • ah , If that is the case , then I apologize for the citing @JohnDoe – Naveen V Mar 15 '23 at 07:50

2 Answers2

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The paper you cite is hard to read as it is written in an unclear way, but it is the case that photons can appear to have a mass when they are travelling through some material like glass, and this may be what the paper is about.

This mass arises because the photon interacts with electrons in the glass and forms a mixed photon/electron state that gets a mass from the electron part of the state. For more on this see my answer to If refraction slows down light, isn't it possible to hold light still? However we should emphasise that this does not mean the photon has a mass. It means only that the mixed photon/electron state has a mass, and that mass depends on the strength of the interaction.

Generations of students have taken the photon energy, $h\nu$, and divided it by $c^2$ to try and calculate a mass for the photon, but the result has no physical meaning. Einstein's original equation $E = mc^2$ applies only to particles that are stationary, and photons are obviously not stationary. For a particle moving with a momentum $p$ the full equation is:

$$ E^2 = p^2c^2 + m^2c^4 $$

where $m$ is the rest mass of the particle. For a photon $m = 0$ and the equation becomes:

$$ E = pc $$

Then setting $E = h\nu = hc/\lambda$ gives us the equation for the momentum:

$$ p = \frac{h}{\lambda} $$

It is tempting to say that $p = mv$ and substitute this to calculate a mass for the photon but the momentum of a photon is not given by this equation - the equation $p=mv$ only applies to massive particles. That is why it is physically meaningless to calculate a photon mass this way.

Strictly speaking the zero rest mass of a photon is just an assumption we make, and it is possible that some experiment one day might discover photons do have a non-zero rest mass. There is an nice review of the experimental evidence on the Science web site. The latest experiments show that if the photon does have a mass it must be less than about $10^{-48}$ kg.

John Rennie
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When nowadays physicists speak of the mass of a particle, they mean the rest mass (the mass of a particle at rest, i.e. when it is not moving), but as photons at rest don't exist, so we can conclude that they don't have a rest mass.

Actually, what you have calculated is not the rest mass of the photon, but the energy (which only differs by a constant scaling factor from what you calculated):

$$ E = \frac{h}{\lambda c} \times c^2 $$

The number which you computed varies with the wave length, so it cannot be the rest mass. Of course to any energy a mass can be attributed according to Einstein's famous formula $E=mc^2$, but this mass -- as already said -- is not the rest mass, in particular as it varies with wave length. Anyway, if we choose the units where $c=1$, there is no difference anymore between energy and the number which was calculated in the post:

$$ E = \frac{h}{\lambda} $$

The choice of using units $c=1$ is often taken to simplify computations by the way.

Frederic Thomas
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