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

And we have the answer

3. Mar 5, 2006

### Pengwuino

Alright i'll try to get to that myself... is there anything special about the equation they wanted me to find?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook