Question

A 5-MeV photon produces a positron-electron pair in a lead shield. If both particles are of equal energy, how far will they travel in the shield?

Answer 1

Gives

Energy (E)=5Mev

We know that

$\lambda=\frac{h}{p}=\frac{hc}{pc}$

We know that

$E^2=p^2c^2+m_0^2c^4$

Then non reletivistic equation

E=pc

$\lambda=\frac{hc}{pc}=\frac{hc}{E}$

$$\lambda=\frac{6.625\times10^{-34}\times3\times10^8}{5\times10^6}=\frac{1.986\times10^{-16}}{5\times1.6\times10^{-13}}nm$$

$\lambda=2.48\times10^{-19}m$