Given a photon dropping from $\lambda_1$ to $\lambda_2$, its energy will drop from $\frac{hc}{(\lambda_1)}$ to $\frac{hc}{(\lambda_2)}$. However, I was wondering if there is any significance in the energy of the change in wavelength itself. For $\Delta \lambda = \lambda_2 – \lambda_1$, this change in wavelength has an energy $\frac{hc}{(\Delta \lambda)}$, yet this value does not correspond at all (as far as I can see) with $\Delta E = \frac{hc}{(\lambda_1)} -\frac{hc}{(\lambda_2)}$. Is this just a subtlety in the math, or is there actually meaning behind the value $\frac{hc}{(\Delta \lambda)}$?
[Physics] Change in Wavelength of a Photon Relation to Energy (specifically Compton Effect)
energyphotonswavelength
Related Solutions
The frequency of light is not a property of many photons but a property of a single photon. (This is also strictly inaccurate, since we should think about a field with ripples in it; we then call a little clump of waves a photon.)
Anyway, let's imagine a photon/wavepacket as having a typical wavelength $\lambda$ given by the peak-peak spacing in this diagram:
This is a single photon, with the colour given by wavelength $\lambda$. Then this whizzes past you at the speed of light, $c$. But then you see the peaks go past at a rate of 1 per $\lambda/c$. Thus there is a sensible definition of frequency as $f=c/\lambda$. Then we let $E=hf$ and carry on happily.
This is all essentially an example of wave-particle duality. If you want to try to understand this better, try reading about the photoelectric effect. The key idea is that there is a wave-like phenomena (light) which we perceive as interacting with the rest of the world in discrete, particle-like events (photons).
Best Answer
If $\Delta \lambda$ is much smaller compared to either $\lambda_1$ or $\lambda_2$(it doesn't really matter, it should be much smaller than both), then we can make the following approximation:
$$|\Delta E|= \left|\Delta \left( \frac {hc}{\lambda}\right)\right|\approx\left|\frac {hc\Delta\lambda}{\lambda^2}\right|$$