How can a red LED work at a voltage of only $1.4{\rm V}$? Red photons have an energy of $1.77{\rm V}$ if we use $\lambda = 0.7\mu\text{m}$ in the relation: $$ E = \hbar \omega = \frac{2\pi\hbar c}{\lambda} \ \ \Rightarrow \ \ E[eV] = \frac{2\pi\hbar c}{\lambda \,e}$$ Even if we assume they are at the edge of the [visible spectrum] at $750{\rm nm}$ that would still be $1.65{\rm V}$. If we assume the energy of each electron in the electric current is converted to photon energy when the electron annihilates with a hole, that would not work.
Is there extra energy added from the thermal motion? Wouldn't that violate thermodynamic principles?! Is there extra voltage from difference in work functions somewhere? But wouldn't that cancel out in a closed circuit?!
[To be precise: normal operating voltage for such a LED is higher, around $1.6\dots1.8 {\rm V}$, but you can easily see that there is still a faint glow at lower voltages than that, only vanishing if it is lowered to about $1.4 {\rm V}$.]
