Emission of photon from quantum jump We know that in an atom,when an electron shifts from one energy level to another energy is emitted which is equal to the difference between the energies of the orbits. And the equation written in books is this $\Delta E=hf$ where $f$ is frequency. My question is why does this consist of a single photon? I have just started learning about light and there i saw that light consists of packets of energies which are photons and a light should consist of a number of photons(maybe infinite,just like water). So,why isn't the equation $\Delta E=nhf$ where $n$ is the number of packets of energy? I think i am wrong so please enlighten me with the proper concept.
 A: Light always has a beginning and an end. Light is produced when excited electrons relax, which means they fall back to a lower energy level in the atom. A photon is emitted in the process. If the excitation takes place over several energy levels, several photons can also be emitted one after the other during the relaxation.
What we see as light is always the sum of the emissions of photons from countless excited particles.
A: The relationship $\Delta E=nhf$ with n=1 or $n>1$ only tells you about the energy balance but not about the probability of the process. For that one has to take into account the absortion cross section $\sigma_{abs}$ and the ingoing photon flux:
$$W^n = \sigma^n_{abs} \cdot F$$
where $W^n$ ($n$ is an index, not an exponent) is the probability rate per second of the considered n-photon process and $F$  is the photon flux. In general the probability rate $W^1$ for the single photon  reaction is rather small, and $W^2$ for a 2-photon reaction is even much smaller, among other things due to $\sigma^2_{abs}<< \sigma^1_{abs}$. Therefore a 2-photon process only rarely happens. Therefore usually only the 1-photon process is considered, in particular if the photon flux is small which occurs most often.
