Why does total internal reflection (TIR) drastically reduces the efficiency of LED? Total internal reflection (TIR) is the primary reason of efficiency loss in LED bulbs. LED chips are made of GaN, which has a much higher refractive index than surrounding materials. Because photons are emitted at all directions, a large percentage of photons are trapped inside due to TIR. These trapped photons are eventually absorbed and become waste heat.

But I don’t understand why photons become waste heat when absorbed, instead of electron-hole pairs that can be reused?
 A: Reusing the photons to produce more electron hole pairs is sometimes called photon recycling, and some photons do get reabsorbed in a productive way, but if the photon energy is already near the bandgap it may not produce an electron hole pair if there is no available state to be excited.
Then if you think about absorption due to defects in the material, and just the width of the natural band edge of the semiconductor, perhaps broadened some from the electric field you have more losses.  These can be non-radiative processes. So even if the material is good and losses are low you lose a small percentage each path-length between bounces. This adds up quickly and exponentially.
So in short, while we teach that the absorption within in the band gap is negligible, that is really in comparison to the very high absorption above the band edge. We teach that the bands are parabolic and sharp, but if you really look at the band edge in detail there are tailing state below the band edge. So with lots of bounces the other absorption adds up and a lot of that is non-radiative and just turns into heat.
A: When photons strike electrons, the electrons get excited to a higher energy level. Then, spontaneously (randomly), they release this excess energy and return back to their original position. In doing so, they release a photon and some heat into the surrounding area.
When there is total internal reflection, this photon doesn't leave the material, and is instead returned back to the same area, and thus it can re-excite another electron. This continuous excitation of electrons and then releasing of photons is what causes the heating up of electrons without the photons being able to escape (thus reducing the efficiency of the LED).
