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When electron making transition from one higher energy state to lower energy state (energy difference $E$) then producing mass less photon with frequency $nu$ where $ Delta E= h nu$(h is Planck constant).We know energy-mas relation $ E=mc^2$.Why not creating some kind particle in this case which has mass m that we could calculate from energy difference of the two states of the electron?Is there any kind critical energy difference$Delta E_c$such that lower than$Delta E_c$ always is creating photon and higher than $Delta E_c$ its value create particle with mass?

You are asking about when a electron transitions from a higher energy level to a lower energy level so I will assume you are asking about relaxation. There are other type of transitions, like the Auger effect.

Now when a atom/electron relaxes, it moves from a higher energy level to a lower energy level as per QM, as you state, and releases a photon.

You are asking why this transition releases a massless gauge boson.

Now the photon is an elementary particle, part of the SM, massless, pointlike. Photons always travel at speed c in vacuum, when measured locally.

You are asking why this gauge boson is massless.

There are two ways to look at this:

1. gauge invariance, if you want to describe a theory with a zero mass vector and relativity, you have to have gauge invariance. And the photon is massless because it is the mediator of the EM force which is long range. It is because of the unbroken U(1) gauge invariance of the EM force.

Though, the gauge fields may become massive via Higgs (W,Z bosons). But that is a short range force.

Why can't gauge bosons have mass?

How does gauge invariance prevent the photon from acquiring a mass?

2. As per SR, anything that travels at the speed of light, cannot have rest mass. This is because it would cost an infinite amount of energy to speed up a massive particle to speed c.

https://en.wikipedia.org/wiki/Photon

https://en.wikipedia.org/wiki/Special_relativity

There are a few reasons why the particle produced needs to be a photon. Aside from conserving energy, we also need to conserve momentum, charge and spin, for example. So you would need to ask what other particle, instead of a photon, could be emitted while satisfying all those conservation requirements.

If you just consider energy and spin conservation, the total amount of energy available in electron transitions in an atom is small, and not enough to make any of the other massive Bosons. To use your terminology, the maximum energy difference in electron transitions, Δ𝐸, is way below the energy Δ𝐸𝑐 you would need to create any of the other known massive particles that satisfy the other conservation requirements.

It's just a conservation of energy equation.
An example of such a case when an electron jumps to a lower shell it emits a photon and this photon itself is captured by an electron in the outer shell and that electron gets emitted from the atom. So in principle till all the conservation laws are satisfied there is always a finite but very less probability for it to happen than just a emission of photon when an electron jumps to a lower shell.

The relation $E = mc^2$ actually only accounts for the rest-mass of a particle, that is, a particle with zero velocity, or equivalently, in its own reference frame. The complete expression is actually $E^2=(mc^2)^2 + (pc)^2$, where $p$ is the momentum and $c$ is the speed of light. Photons have zero mass, but they have non-zero momentum. When $m=0,$ the expression reduces to $E^2 = (pc)^2$ or $E = pc$. By the de-Broglie relation, $p=h/lambda$, so $E=hc/lambda=hnu$, as you have suggested.