Question 705075

SCQHARD

The $\Delta^{++}$ can be produced by colliding a pion beam onto a $H_2$ target, in a reaction $\pi^+ + p \rightarrow \Delta^{++} \rightarrow \pi^+ + p$ . In the rest frame of $\Delta^{++}$ , the energy and momentum of the pion in the final state (in MeV) are closest to (assume $c = 1$ , and $m_{\pi} \approx 140 \text{MeV}$ , $m_{p} \approx 1 \text{GeV}$ , $m_{\Delta^{++}} \approx 1.2 \text{GeV}$ )

(A)

210,156

(B)

230,182

(C)

175,105

(D)

190,130

Detailed Solution

In the rest frame of $\Delta^{++}$ , the total energy is $E_{\Delta} = 1200 \text{MeV}$ . By conservation of energy, $E_{\pi} + E_p = 1200$ . Using energy conservation in particle decay, $E_{\pi} = \frac{m_{\Delta}^2 + m_{\pi}^2 - m_p^2}{2m_{\Delta}} = \frac{1200^2 + 140^2 - 1000^2}{2(1200)} \approx 191.5 \text{MeV}$ . Momentum $p = \sqrt{E_{\pi}^2 - m_{\pi}^2} = \sqrt{191.5^2 - 140^2} \approx 130.7 \text{MeV}$ . Closest option is 190,130.

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