Question

Some particle accelerators allow protons$1p+2$and antiprotons$1p-2$to circulate at equal speeds in opposite directions in a device called a storage ring. The particle beams cross each other at various points to cause$p^{+}+p^{-}$collisions. In one collision, the outcome is$p^{+}+p^{-}\to e^{+}+e^{-}+\gamma +\gamma$, where g represents a high-energy gamma-ray photon. The electron and positron are ejected from the collision at$0.9999995c$and the gamma-ray photon wavelengths are found to be$1.0\times {10}^{-6}nm$. What were the proton and antiproton speeds, as a fraction of $c$, prior to the collision?

Short answer

The proton and antiproton speeds is $0.845c$as fraction of $c$, prior to the collision

Step-by-step solution

Given Information

We need to find the proton and antiproton speeds, as a fraction of c, prior to the collision.

Simplify

The energy of each electron and positron is $E_e=\frac{m_e{c}^2}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}$. Now, $m_e=9.1\times {10}^{-31}kg$and $c=3\times {10}^{8}m/s$ $v=0.9999995c$. Putting the value in the formula:

$E_e=8.19\times {10}^{-11}J$

Now, the energy of the gamma-ray is $E_{\gamma }=\frac{hc}{\lambda }$.

We have $\lambda =1.0\times {10}^{-6}nm$and $h=6.626\times {10}^{-34}k{g}^{2}m/s.$

$E_{\gamma }=1.9\times {10}^{-10}$

Therefore, the total energy is

$E_{t}ot=E_e+E_{\gamma }=2.81\times {10}^{-10}$

Now the energy of each proton is

$E_p=\frac{m_p{c}^2}{\sqrt{1-{\displaystyle \raisebox{1ex}{${v^2}_p$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}$

We have $m_p=1.67\times {10}^{-27}kg$.

$2.81\times {10}^{-10}=\frac{1.67\times {10}^{-27}\times {\left(3\times {10}^8\right)}^2}{\sqrt{1-{\displaystyle \raisebox{1ex}{${v^2}_p$}\!\left/ \!\raisebox{-1ex}{${\left(3\times {10}^8\right)}^2$}\right.}}}$

Solving the above question we get:

$v_p=2.535\times {10}^{8}m/s=\frac{2.535\times {10}^8}{3\times {10}^8}c=0.845c$

So the speed of each proton is$0.845c$