Chapter 32, Problem 53GP

(a)

To determine

The $Q$ for the $p+p\to p+p+\pi \text{'}$

Answer to Problem 53GP

  $Q=-135\text{ MeV}$

Explanation of Solution

Given:

  $p+p\to p+p+\pi \text{'}$

  $m_{\pi }=135\frac{\text{MeV}}{c^2}$

Formula used:

  $Q=\left(2m_p{c}^2-2m_p{c}^2-m_{\pi }{c}^2\right)$

Where,

  $m_p$ = mass of proton

  $m_{\pi }$ = mass of pion

  $c$ = speed of light

Calculation:

Using the formula, the $Q$ -value is calculated as,

  $Q=\left(2m_p{c}^2-2m_p{c}^2-m_{{\pi }^0}{c}^2\right)$

  $\begin{array}{l}Q=-m_{{\pi }^0}{c}^2\\ Q=-\left(\frac{135\text{ MeV}}{c^2}\right)c^2\\ Q=-135\text{ MeV}\end{array}$

(b)

To determine

The $Q$ - value of $p+p\to p+n+{\pi }^{+}$

Answer to Problem 53GP

  $Q=-140.9\text{ MeV}$

Explanation of Solution

Given:

  $p+p\to p+n+{\pi }^{+}$

  $m_{{\pi }^{+}}=139.6\frac{\text{MeV}}{c^2}$

Formula used:

  $Q=\left(2m_p{c}^2-m_p{c}^2-m_n{c}^2-m_{{\pi }^{+}}{c}^2\right)$

Calculation:

Using the formula, the $Q$ -value is calculated as,

  $\begin{array}{l}Q=\left(2m_p{c}^2-m_p{c}^2-m_n{c}^2-m_{{\pi }^{+}}{c}^2\right)\\ Q=\left(2\times 938.3-938.3-939.6-139.6\right)\text{MeV}\\ Q=-140.9\text{ MeV}\end{array}$