# Why doesn't photon lose energy over distance?

In an atom, ( or even an oscillating circuit) When an electron falls from higher energy level to a lower one, it gives a photon of energy $hf$. On other hand it can be considered as an accelerated-decelerated charged particle that make an electromagnetic radiation which moves vertical to the path of acceleration. But in first explanation we have a photon that goes into the vacuum and does not lose energy over distance but in latter explanation we have an electromagnetic wave that diminishes by the ratio of $1/r^2$.
How do you interpret this discrepancy? I am sure I am making a mistake somewhere.

## 2 Answers

A classical electromagnetic wave can be thought of as a very large number of photons. The energy carried by the wave is then given by the number of photons multiplied by the energy per photon (assuming a monochromatic source for simplicity). As the photons travel, simply by geometry the number of photons per unit area will decrease as $\frac {1}{r^2}$, while the energy per photon is unchanged. So the energy per unit area in the wave decreases as $\frac {1}{r^2}$.

Photon is no special in this regard. Anything that moves through space, does not lose energy. You throw a rock in space by giving it a speed of V. It can travel millions or billions of miles through space and still it has same kinetic energy that was imparted to it in the beginning. Just moving through space is no reason to lose energy. So, that also means no energy is consumed in moving something from point A to point B in space. The energy you impart in getting it moving, can be fully recovered at destination provided you have a mechanism to do so. Only intensity decreases with distance because the photons are spread more thinly at greater distance from the source.