smc-3704-30-Nuclear Transformation

The simplified model below shows the reactants and products of a proton-antiproton reaction which produces three particles called pions, each having a different charge.

\(\text{p}+\overline{\text{p}} \rightarrow \pi^{+}+\pi^0+\pi^{-}\)

There are no other products in this process, which involves only the rearrangement of quarks. No electromagnetic radiation is produced. Assume that the initial kinetic energy of the proton and antiproton is negligible.

Protons consist of two up quarks \(\text{(u)}\) and a down quark \(\text{(d)}\) . Antiprotons consist of two up antiquarks \((\overline{\text{u}})\) and a down antiquark \((\overline{\text{d}})\). Each of the pions consists of two quarks.

The following tables provide information about hadrons and quarks.

Table 1: Hadron Information

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \quad \ \ \textit{Particle} & \ \ \textit{Rest mass} \ \ & \quad \textit{Charge} \quad \\
& \left(\text{MeV/c}^2\right)&\\
\hline
\rule{0pt}{2.5ex} \text {proton (p)} \rule[-1ex]{0pt}{0pt} & 940 & +1 \\
\hline
\rule{0pt}{2.5ex} \text {antiproton}(\overline{\text{p}}) \rule[-1ex]{0pt}{0pt} & 940 & -1 \\
\hline
\rule{0pt}{2.5ex} \text {neutral pion }\left(\pi^0\right) \rule[-1ex]{0pt}{0pt} & 140 & \text{zero} \\
\hline
\rule{0pt}{2.5ex} \text{positive pion }\left(\pi^{+}\right) \rule[-1ex]{0pt}{0pt} & 140 & +1 \\
\hline
\rule{0pt}{2.5ex}\text {negative pion }\left(\pi^{-}\right) \rule[-1ex]{0pt}{0pt} & 140 & -1\\
\hline
\end{array}

Table 2: Quark charges

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \quad \quad \ \ \textit{Particle} \rule[-1ex]{0pt}{0pt} & \quad \textit{Charge} \quad \\
\hline
\rule{0pt}{2.5ex} \text {down quark (d)} \rule[-1ex]{0pt}{0pt} & -\dfrac{1}{3} \\
\hline
\rule{0pt}{2.5ex} \text {up quark (u)} \rule[-1ex]{0pt}{0pt} & +\dfrac{2}{3}\\
\hline
\rule{0pt}{2.5ex} \text {down antiquark}(\overline{\text{d}}) \rule[-1ex]{0pt}{0pt} & +\dfrac{1}{3}\\
\hline
\rule{0pt}{2.5ex} \text{up antiquark }(\overline{\text{u}}) \rule[-1ex]{0pt}{0pt} & -\dfrac{2}{3} \\
\hline
\end{array}

Identify the quarks present in the \(\pi^{-}, \pi^{+}\)and the \(\pi^0\) particles. (2 marks)

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The energy released in the reaction is shared equally between the pions.
Calculate the energy released per pion in this reaction. (2 marks)

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Calculation of the pions' velocities using classical physics predicts that each pion has a velocity, relative to the point at which the proton-antiproton reaction occurred, which exceeds 3 × 10\(^8\) m s\(^{-1}\).
Explain the problem with this prediction and how it can be resolved. (3 marks)

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a. \(\pi^{-}:\ \overline{\text{u}}\text{d}\)

\(\pi^{+}:\ \text{u}\overline{\text{d}}\)

\(\pi^{0}:\ \text{u}\overline{\text{d}}\)

b. \(\text{Initial mass}\ = 940 + 940 = 1880\ \text{MeV/c}^{2} \)

\(\text{Final mass}\ = 3 \times 140 = 420\ \text{MeV/c}^{2} \)

\(\Delta \text{Mass (per pion)}\ = \dfrac{1460}{3} = 487\ \text{MeV/c}^{2} \)

\(\text{Using}\ \ E=mc^2:\)

\(\text{Energy released (per pion)}\ = 487\ \text{MeV}\)

c. Problem with prediction:

→ The calculation shows the pions moving faster than light speed (3 × 10\(^{8}\) m/s), which can’t happen in reality.

Resolving the problem:

→ Since these pions are moving at extremely high speeds, we need to account for relativity.

→ Relativity means that pions’ mass actually increases as they get faster, which prevents them from ever reaching light speed.

→ Part of the energy given to the pions goes into increasing their mass rather than just increasing their velocity.

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