Products Rewards
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY
HELP DESK

Need Help? We got you covered.

## Q. 14.10

(a) How much energy was available in the center of mass for the experiment of Segrè and Chamberlain, who used 6.4-GeV protons on a fixed proton target to produce antiprotons in the reaction given here?

$p+p \rightarrow p+p+p+\bar{p}$

(b) How much beam energy was necessary to produce the antiprotons? (c) How much energy is available for a similar reaction with 1-TeV protons from the Tevatron on a fixed proton target?

Strategy (a) We can use Equation (14.10) to calculate the energy available in the center of mass. (b) The threshold kinetic energy can be found by using Equation (14.11). We use the Q value defined in Equation (13.7). (c) We use Equation (14.10) to determine the center-of-mass energy available for the reaction for the Tevatron.

$Q=M_{x} c^{2}+M_{X} c^{2}-\left(M_{y} c^{2}+M_{Y} c^{2}\right)=K_{y}+K_{Y}-K_{x}$ (13.7)

$E_{ cm }=\sqrt{\left(m_{1} c^{2}+m_{2} c^{2}\right)^{2}+2 m_{2} c^{2} K}$ (14.10)

$K_{ th }=(-Q) \frac{\text { total masses involved in reaction }}{2 m_{2}}$ (14.11)

## Verified Solution

(a) If we use Equation (14.10) and the rest energy of the proton (938 MeV), we have

\begin{aligned}E_{ cm } &=\sqrt{[2(0.938 GeV )]^{2}+2(0.938 GeV )(6.4 GeV )} \\&=3.94 GeV\end{aligned}

The total mass of the reaction products is $4\left(0.938 GeV / c^{2}\right)=3.75 GeV / c^{2}$, so the Bevatron was constructed with just enough energy to create the four particles in the final state.

(b) Equation (13.7) gives the Q value:

\begin{aligned}Q &=(\text { Initial mass energies })-(\text { final mass energies }) \\&=2 m_{p} c^{2}-4 m_{p} c^{2}=-2 m_{p} c^{2} \\&=-2(0.938 GeV )=-1.88 GeV\end{aligned}

If we insert this value into Equation (14.11), we obtain

$K_{ th }=(-Q) \frac{6 m_{p}}{2 m_{p}}=(1.88 GeV )(3)=5.6 GeV$

The kinetic energy of 6.4 GeV was clearly enough to initiate the reaction, which is consistent with our result in (a).

(c) Now we insert the kinetic energy of 1 TeV into Equation (14.10) to determine

\begin{aligned}E_{ cm } &=\sqrt{[2(0.938 GeV )]^{2}+2(0.938 GeV )(1000 GeV )} \\&=43 GeV\end{aligned}

Because of the conservation of momentum requirement, there was a tremendous reduction in the available energy for reactions for the 1000-GeV proton beam from the Tevatron.