# Percentage increase in wavelength leads to percentage decrease in energy. (Compton effect)

In this question:

What percentage increase in wavelength leads to a 75% loss of photon energy in a photon-free electron collision?

My understanding of this question is that $$0.25\times E_i = E_f$$, and as $$E=\frac{hc}{\lambda}$$, $$\lambda_i=0.25\times \lambda_f$$, so if you increase $$\lambda$$ by 3 times itself, it would lead to a 75% loss.

But I was instead told that I shouldn't use that method and should instead use $$\Delta E = 0.75E_i$$. And $$\frac{\Delta E}{E_i}= \frac{\Delta\lambda}{\Delta\lambda+\lambda_i},$$ resulting in $$\frac{\Delta\lambda}{\lambda_i}=\frac{\Delta E/E_i}{1 - \Delta E/E_i}$$ i.e. $$0.75/(1-0.75)=3$$.

This may simply be my math failing me or may represent a greater problem so i would greatly appreciate any insights as to why the second method would be preferable over the first.

• Hi J Doe and welcome to the site! You're right that we don't really give homework help here but this question is fine because you're not actually asking for help with the homework problem, you're asking a conceptual question which just happens to have come up in the context of a homework problem. – David Z Oct 7 '18 at 2:49
• Okay, that's excellent, I want to make sure I follow the rules that have been set. Also, thanks @David for the help editing, the post is much improved after your work. – J Doe Oct 7 '18 at 2:54