Chapter 32, Problem 25P

(a)

To determine

TheQ -value for the decay $\pi °\to p+{\pi }^{-}$ .

Answer to Problem 25P

Solution:

The Q -value for the decay $\pi °\to p+{\pi }^{-}$ is $37.8\text{MeV}$ .

Explanation of Solution

Explanation:Given info.

  $\pi °\to p+{\pi }^{-}$

Formula used:

The amount of energy released during the reaction, $Q=m{c}^2$

Calculation:

  $\pi °\to p+{\pi }^{-}$

  $\begin{array}{c}Q=\left[m_{\pi °}-m_p-m_{{\pi }^{-}}\right]c^2\end{array}$

   $\begin{array}{c}=m_{\pi °}{c}^2-m_p{c}^2-m_{{\pi }^{-}}{c}^2\end{array}$

Substitute

  $\begin{array}{l}{m}_{\pi °}{c}^2=1115.7\text{MeV}\end{array}$

   $\begin{array}{c}{m}_p{c}^2=938.3\text{MeV}\end{array}$

  $\begin{array}{c}{m}_{{\pi }^{-}}{c}^2=139.6\text{MeV }\end{array}$

    $\begin{array}{c}Q=\left[1115.7-938.3-139.6\right]\text{MeV}\end{array}$

     $\begin{array}{c}=37.8\text{MeV}\end{array}$

Conclusion:

The Q -value for the decay $\pi °\to p+{\pi }^{-}$ is $37.\text{8}\text{MeV}$ .

(b)

To determine

The kinetic energy of the p and $p\text{ and }{\pi }^{-}$ assuming the $\pi °$ decays from rest.

Answer to Problem 25P

Solution:

The kinetic energy of the $p\text{ and }{\pi }^{-}$ are $5.4\text{MeV and }32.4\text{MeV}$ .

Explanation of Solution

Explanation:Given info.

Given particle are $p\text{ and }{\pi }^{-}$

Formula used:

Let $K.E_p\text{ and }K.E_{{\pi }^{-}}$ are the kinetic energy of the proton and pion.

  $Q=K.E_p+K.E_{{\pi }^{-}}$

Apply the principle of conservation of momentum.

  $P_p=P_{{\pi }^{-}}=P$ And

    $K.E_p=E_p-m_p{c}^2$

Calculation:

  $\begin{array}{c}K.E_p=\sqrt{{\left(P_{p}c\right)}^2+{\left(m_p{c}^2\right)}^2}-m_p{c}^2\end{array}$

     $\begin{array}{c}=\sqrt{{\left(Pc\right)}^2+{\left(m_p{c}^2\right)}^2}-m_p{c}^2\end{array}$

And $K.E_{\pi }=\sqrt{{\left(Pc\right)}^2+{\left(m_{{\pi }^{-}}{c}^2\right)}^2}-m_{{\pi }^{-}}{c}^2$

   $\begin{array}{c}Q=\sqrt{{\left(Pc\right)}^2+{\left(m_p{c}^2\right)}^2}-m_p{c}^2+\sqrt{{\left(Pc\right)}^2+{\left(m_{{\pi }^{-}}{c}^2\right)}^2}-m_{{\pi }^{-}}{c}^2\end{array}$

  $\begin{array}{c}37.8=\sqrt{{\left(Pc\right)}^2+{\left(938.3\right)}^2}-938.3+\sqrt{{\left(Pc\right)}^2+{\left(139.6\right)}^2}-139.6\end{array}$

  $\begin{array}{c}1115.7=\sqrt{{\left(Pc\right)}^2+{\left(938.3\right)}^2}+\sqrt{{\left(Pc\right)}^2+{\left(139.6\right)}^2}\end{array}$

Let

  $\begin{array}{l}Pc=x,938.3\text{ be a, 139}\text{.6 be b and 115}\text{.7 be c}\text{.}\end{array}$

  $\begin{array}{c}\text{c}=\sqrt{x^2+a^2}+\sqrt{x^2+b^2}\end{array}$

Squaring both sides,

          $\begin{array}{l}{c}^2=x^2+a^2+x^2+b^2+2\sqrt{\left(x^2+a^2\right)\left(x^2+b^2\right)}\end{array}$

    $\begin{array}{c}{c}^2-\left(x^2+a^2\right)=2x^2+2\sqrt{x^4+\left(x^2+a^2\right)x^2+a^2{b}^2}\end{array}$

$\begin{array}{c}\left[c^2-\left(x^2+a^2\right)-2x^2\right]=2\sqrt{x^4+\left(x^2+a^2\right)x^2+a^2{b}^2}\end{array}$

Squaring both sides,

  $\begin{array}{l}{\left[c^2-\left(x^2+a^2\right)\right]}^2+{\left(2x^2\right)}^2-2\left(c^2-\left(x^2+a^2\right)\right).2x^2=4\left[x^4+\left(x^2+a^2\right)x^2+a^2{b}^2\right]\end{array}$

   $\begin{array}{c}{\left[c^2-\left(x^2+a^2\right)\right]}^2+{\left(2x^2\right)}^2-4x^2\left[c^2-\left(x^2+a^2\right)\right]=4x^4\left(x^2+a^2\right)x^2+4a^2{b}^2\end{array}$

       $\begin{array}{c}{\left[c^2-\left(x^2+a^2\right)\right]}^2-4x^2{c}^2+4x^2\left(x^2+a^2\right)=4\left(x^2+a^2\right)x^2+4a^2{b}^2\end{array}$

              $\begin{array}{c}{\left[c^2-\left(x^2+a^2\right)\right]}^2-4x^2{c}^2=4a^2{b}^2\end{array}$

  $\begin{array}{c}{x}^2=\frac{{\left[c^2-\left(x^2+a^2\right)\right]}^2-4a^2{b}^2}{4c^2}\end{array}$

  $x=\frac{{\left[\left(c^2-\left(a^2+b^2\right)\right)-4{\left(ab\right)}^2\right]}^{\frac{1}{2}}}{2c}$

   $\begin{array}{c}=\frac{{\left({1115.7}^2-\left({938.3}^2+{139.6}^2\right)-4{\left(938.3\times 139.6\right)}^2\right)}^{\frac{1}{2}}}{2\times 1115.7}\end{array}$

  $\begin{array}{c}x=100.5\text{MeV}\end{array}$

Therefore, $Pc=100.5MeV$

And now,

  $\begin{array}{c}K.E_p=\sqrt{{\left(Pc\right)}^2+{\left(m_p{c}^2\right)}^2}-m_p{c}^2\end{array}$

     $\begin{array}{c}=\sqrt{{\left(100.5\right)}^2+{938.3}^2}-938.3\text{MeV}\end{array}$

     $\begin{array}{c}=5.4\text{MeV}\end{array}$

  $\begin{array}{c}K.E_{{\pi }^{-}}=\sqrt{{\left(Pc\right)}^2+{\left(m_{{\pi }^{-}}{c}^2\right)}^2}-m_{{\pi }^{-}}{c}^2\end{array}$

     $\begin{array}{c}=\sqrt{{\left(100.5\right)}^2+{139.6}^2}-139.\text{6MeV}\end{array}$

     $\begin{array}{c}=32.4\text{MeV}\end{array}$

Conclusion:

The K.E of $p\text{ and }{\pi }^{-}$ are $5.4\text{MeV}$ and $32.4\text{MeV}$ .