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for the following question:

Consider a pion ($π^\pm$) scattering on a target. What pion momenta are needed to observe structures of the size of an atom ($10^{-10}$ m), an atomic nucleus ($10^{−14}$ m), or the proton ($10^{−15}$ m)? For each case, consider if you need a relativistic treatment and explain your reasoning.

I know the Broglie wavelength is $h/p$. But for the $10^{-14}$ m and $10^{-15}$ m I need to treat it relativistically, how do I do this?

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The de Broglie relation $$\lambda=\frac{h}{p} \tag{1}$$ is already relativistically correct. So you only need simple math to find the momentum $p$.

May be you confused it with the relation between wavelength $\lambda$, velocity $v$ and rest mass $m$ $$\lambda=\frac{h\sqrt{1-\frac{v^2}{c^2}}}{mv}. \tag{2}$$ You can get this equation by combining the de Broglie relation (1) with the relativistic momentum $p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$ To solve (2) for the velocity $v$ you would indeed need some more complicated math.

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  • $\begingroup$ But for instance, if I use 10^-14 m, I get a momentum that leads to a velocity that is almost equal to c, if I use 10^-15m, I get a momentum that leads to a speed that is more then the speed of light. So something goes wrong here $\endgroup$ Commented Nov 17 at 11:31
  • $\begingroup$ @dutchrunner No, by using $$p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$$ you always get a velocity $v$ smaller than $c$., regardless how big $p$ may be, $\endgroup$ Commented Nov 17 at 11:36

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