Question

Figure 44-14 shows part of the experimental arrangement in which antiprotons were discovered in the 1950s. A beam of 6.2 GeV protons emerged from a particle accelerator and collided with nuclei in a copper target. According to theoretical predictions at the time, collisions between protons in the beam and the protons and neutrons in those nuclei should produce antiprotons via the reactions

However, even if these reactions did occur, they would be rare compared to the reactions

Thus, most of the particles produced by the collisions between the 6.2 GeV protons and the copper target were pions.

To prove that antiprotons exist and were produced by some limited number of the collisions, particles leaving the target were sent into a series of magnetic fields and detectors as shown in Fig. 44-14. The first magnetic field (M1) curved the path of any charged particle passing through it; moreover, the field was arranged so that the only particles that emerged from it to reach the second magnetic field (Q1) had to be negatively charged (either a ) and have a momentum of . Field Q1 was a special type of magnetic field (a quadrapole field) that focused the particles reaching it into a beam, allowing them to pass through a hole in thick shielding to a scintillation counter S1. The passage of a charged particle through the counter triggered a signal, with each signal indicating the passage of either a or (presumably) a 1.19 GeV/c p¯. After being refocused by magnetic field Q2, the particles were directed by magnetic field M2 through a second scintillation counter S2 and then through two Cerenkov counters C1 and C2. These latter detectors can be manufactured so that they send a signal only when the particle passing through them is moving with a speed that falls within a certain range. In the experiment, a particle with a speed greater than 0.79c would trigger C1 and a particle with a speed between 0.75c and 0.78c would trigger C2.

There were then two ways to distinguish the predicted rare antiprotons from the abundant negative pions. Both ways involved the fact that the speed of a 1.19 GeV/c p¯ differs from that of a 1.19 GeV/c

: (1) According to calculations, a p¯ would trigger one of the Cerenkov counters and a would trigger the other. (2) The time interval t between signals from S1 and S2, which were separated by 12 m, would have one value for a p¯ and another value for a${\mathbf{\pi }}^{\mathbf{-}}$. Thus, if the correct Cerenkov counter were triggered and the time interval $\mathbf{△}\mathbf{t}$had the correct value, the experiment would prove the existence of antiprotons.

What is the speed of (a) an antiproton with a momentum of 1.19 GeV/c and (b) a negative pion with that same momentum? (The speed of an antiproton through the Cerenkov detectors would actually be slightly less than calculated here because the antiproton would lose a little energy within the detectors.) Which Cerenkov detector was triggered by (c) an antiproton and (d) a negative pion? What time interval $\mathbf{∆}\mathbf{t}$indicated the passage of (e) an antiproton and (f) a negative pion? [Problem adapted from O. Chamberlain, E. Segrè, C.Wiegand, and T.Ypsilantis,“Observation of Antiprotons,”Physical Review, Vol. 100, pp. 947–950 (1955).]

Short answer

(a)The speed is $\mathrm{v}=0.785\mathrm{c}$.

(b)The speed is $\mathrm{v}=0.993\mathrm{c}$.

(c) Because the speed of the antiproton is about 0.78c and it is not more than or over the 0.79c, the antiproton will trigger C2.

(d) Because the speed of the negative pions exceeds 0.79c, the negative proton will trigger the C1.

(e) The time interval is $\mathrm{\Delta t}=5.1\times {10}^{-8}\text{s}=\text{51}\text{ns}$.

(f) The time interval is $\mathrm{\Delta t}=4\times {10}^{-8}\text{s}=40\text{ns}$.

Step-by-step solution

Determine the concept

The speed of any object is known as the distance travel in a given time interval and themomentum p of object can be define as the product of the mass of the object and the speed of the object.

(a) Determine the speed of an antiproton with a momentum of 1.19 GeV/c

The relative relation between the speed and the momentum will be

$$\begin{gathered} \mathrm{p}=\mathrm{\gamma mv} \\ =\frac{\mathrm{mv}}{\sqrt{1-{\left(\mathrm{v}/\mathrm{c}\right)}^2}} \\ \frac{\mathrm{v}}{\mathrm{c}}=\sqrt{1-\frac{1}{\left(\mathrm{pc}/{\mathrm{mc}}^2\right)+1}} \end{gathered}$$

For the antiproton the values are as follows:

$$\begin{gathered} {\mathrm{mc}}^2=938.3\text{MeV} \\ \mathrm{pc}=1.19\text{GeV}=1190\text{MeV} \end{gathered}$$

Solve for the velocity as:

$$\begin{gathered} \mathrm{v}=\mathrm{c}\sqrt{1-\frac{1}{\left(1190\mathrm{Mev}+938.3\mathrm{Mev}\right)+1}} \\ \mathrm{v}=0.785\mathrm{c} \end{gathered}$$

(b) Determine the speed of a negative pion with that same momentum  

For the negative pion the values are as follows:

$$\begin{gathered} {\mathrm{mc}}^2=193.6\text{MeV} \\ \mathrm{pc}=1.19\mathrm{GeV}=1190\text{MeV} \end{gathered}$$

Determine the speed as:

$$\begin{gathered} \mathrm{v}=\mathrm{c}\sqrt{1-\frac{1}{(1190\mathrm{Mev}+193.6\mathrm{Mev})+1}} \\ \mathrm{v}=0.993\mathrm{c} \end{gathered}$$

(c) Determine the detector that was triggered by an antiproton

Because the speed of the antiproton is about 0.78c and it is not more than or over the 0.79c, the antiproton will trigger C2.

(d) Determine detector that is triggered by a negative pion and time interval that indicated the passage

Because the speed of the negative pions exceeds 0.79c, the negative proton will trigger the C1.

(e) Determine the time interval indicated by an antiproton

For an antiproton, d distance, v velocity and delta t is time interval.

$\mathrm{\Delta t}=\frac{\mathrm{d}}{\mathrm{v}},\mathrm{d}=1\text{m}$

Solve for the time interval as:

$$\begin{gathered} △\mathrm{t}=\frac{1}{0.9758\left(2.998\times {10}^8\mathrm{m}/\mathrm{s}\right)} \\ △\mathrm{t}=5.1\times {10}^8 \\ △\mathrm{t}=51\mathrm{ns} \end{gathered}$$

(f) Determine the time interval indicated by negative pion

For the negative pion the condition will be

$\mathrm{\Delta t}=\frac{\mathrm{d}}{\mathrm{v}},\mathrm{d}=12\text{m}$

Solve for the time interval as:

$$\begin{gathered} △\mathrm{t}=\frac{12}{0.993\left(2.998\times {10}^{8}m/s\right)} \\ △\mathrm{t}=4.0\times {10}^{-8} \\ △\mathrm{t}=40\mathrm{ns} \end{gathered}$$