Why do hot objects prefer to emit photons over electrons ? Is there electron-positron annihilation ? If so , why ? Im confused by this.
[Physics] Why do hot objects prefer to emit photons over electrons ? Is there electron-positron annihilation
electromagnetismparticle-physicsthermodynamicsvisible-light
Related Solutions
It's possible, just very unlikely. You can get a clue of the relevant probabilities by looking at the Feynman diagrams for different kinds of $e^+e^-$ annihilation. Here's $e^+e^-\to\gamma\gamma$:
The probability of this occurring (actually, the cross section) is proportional to a factor of $g_\text{EM}$ for each vertex. $g_\text{EM}$ is the electromagnetic coupling, which has a value of about 0.3. So the probability of the entire process can be represented as proportional to $\alpha_\text{EM} = \frac{g_\text{EM}^2}{4\pi} \approx \frac{1}{137}$.
For neutrino production, on the other hand, the simplest Feynman diagram is this:
The probability of this is proportional to two factors of the weak coupling, $g_\text{weak}$, and $\alpha_\text{weak} = \frac{g_\text{weak}^2}{4\pi} \approx 10^{-6}$ (source). So this process is on the order of 10000 times less likely than the annihilation into photons. (In fact, it's actually even less likely than that, because at low energies, as akhmeteli pointed out, the probability is further suppressed by a factor of $m_W^{-2}$, where $m_W$ is the relatively large mass of the W boson.)
Gravity is an even weaker force, so we would expect the corresponding diagram for annihilation into gravitons to be much less probable. You can estimate that $\alpha_\text{gravity} \approx 10^{-39}$. But in this case, it's not even clear how well Feynman diagrams describe the process at all, since we don't have a proper quantum theory of gravity.
Electron positron annihilations can give mu and tau neutrinos as well as electron neutrinos. For a calculation of the probabilities see for example Mu and Tau Neutrino Thermalization and Production in Supernovae: Processes and Timescales.
You might also be interested to read DavidZ's answer to Why does electron-positron annihilation prefer to emit photons?.
Best Answer
The energy levels in any object are quantized. The ground state of an electron is the lowest energy. From there electrons can have many higher energy levels, the highest being that which allows them to completely escape the nuclear attraction, i.e., get emitted.
So hot objects can emit electrons, but the probability that any single electron will have that much energy just due to heating the object is lower than electrons going up a few levels (due to gaining thermal energy) and then jumping back and emitting photons. Very hot objects can become plasma (electrons are emitted and you are left with ions). There are of course other ways to achieve a plasma (e.g., dielectric breakdown discharge) which are not equilibrium thermal excitation. But that is beyond scope of this question.