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Chapter 36: Q. 75 (page 1062)

Some particle accelerators allow protons1p+2and antiprotons1p-2to circulate at equal speeds in opposite directions in a device called a storage ring. The particle beams cross each other at various points to causep++p-collisions. In one collision, the outcome isp++p-鈫�e++e-+纬+纬, where g represents a high-energy gamma-ray photon. The electron and positron are ejected from the collision at0.9999995cand the gamma-ray photon wavelengths are found to be1.0脳10-6nm. What were the proton and antiproton speeds, as a fraction of c, prior to the collision?

Short Answer

Expert verified

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

Step by step solution

01

Given Information

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

02

Simplify

The energy of each electron and positron is Ee=mec21-v2c2. Now, me=9.1脳10-31kgand c=3脳108m/s v=0.9999995c. Putting the value in the formula:

Ee=8.19脳10-11J

Now, the energy of the gamma-ray is E纬=hc位.

We have 位=1.0脳10-6nmand h=6.626脳10-34kg2m/s.

E纬=1.9脳10-10

Therefore, the total energy is

Etot=Ee+E纬=2.81脳10-10

Now the energy of each proton is

Ep=mpc21-v2pc2

We have mp=1.67脳10-27kg.

2.81脳10-10=1.67脳10-27脳3脳10821-v2p3脳1082

Solving the above question we get:

vp=2.535脳108m/s=2.535脳1083脳108c=0.845c

So the speed of each proton is0.845c

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Most popular questions from this chapter

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