# Homework Help: A KE derivation from Compton Effect

1. Mar 5, 2006

### Pengwuino

I have an odd problem here. I need to show that:

$$KE = \frac{{\frac{{\Delta \lambda }}{\lambda }}}{{1 + (\frac{{\Delta \lambda }}{\lambda })}}hf$$

I've basically derived $$KE = \frac{{hc}}{{\lambda _o }} - \frac{{hc}}{{\lambda '}}$$ down to…

$$KE(\frac{{\lambda '}}{{\lambda _o ^2 }}) = (\frac{{\Delta \lambda }}{{\lambda _o }})hf$$

but I'm not sure how I can turn that $$\frac{{\lambda '}}{{\lambda _o ^2 }}$$ into a $$1 + (\frac{{\Delta \lambda }}{{\lambda _o }})$$

Can anyone help?

Last edited: Mar 5, 2006
2. Mar 5, 2006

### phucnv87

$$\Delta E=\frac{hc}{\lambda_0}(1-\frac{\lambda_0}{\lambda})$$

$$1-\frac{\lambda_0}{\lambda}=\frac{\lambda-\lambda_0}{\lambda}=\frac{\frac{\lambda-\lambda_0}{\lambda_0}}{1+\frac{\lambda-\lambda_0}{\lambda_0}}$$

Consider $$\frac{\Delta \lambda}{\lambda_0}=\frac{\lambda-\lambda_0}{\lambda_0}$$