Chapter 44, Problem 13P

(a)

To determine

The Q value of the decay of ${\Lambda }^0$ particle.

Answer to Problem 13P

The Q value of the decay reaction is $37.7\text{MeV}$.

Explanation of Solution

The ${\Lambda }^0$ particle decay into a ${\pi }^{-}$ meson. Write the decay reaction, keeping in mind the required conservation laws.

    ${\Lambda }^0\to \text{p}+{\pi }^{-}$

Write the formula for the Q value of the above decay reaction.

    $Q=m_{\Lambda }{c}^2-m_{\text{P}}{c}^2-m_{\pi }{c}^2$

Here, $Q$ is the Q value of the decay reaction, $m_{\Lambda }{c}^2$ is the rest energy of ${\Lambda }^0$, $m_{\text{P}}{c}^2$ is the rest energy of proton, and $m_{\pi }{c}^2$ is the rest mass of the ${\pi }^{-}$.

Conclusion:

Substitute $1115.6\text{MeV}$ for $m_{\Lambda }{c}^2$, $938.3\text{MeV}$ for $m_{\text{P}}{c}^2$, $139.6\text{MeV}$ for $m_{\pi }{c}^2$.

    $\begin{array}{c}Q=1115.6\text{MeV}-938.3\text{MeV}-139.6\text{MeV}\\ =37.7\text{MeV}\end{array}$

The Q value of the decay reaction is $37.7\text{MeV}$.

(b)

To determine

The total kinetic energy shared by the proton and the ${\pi }^{-}$ meson.

Answer to Problem 13P

The kinetic energy of the products is $37.7\text{MeV}$.

Explanation of Solution

The Q value is the difference in the kinetic energy of the reactants and the products. The kinetic energy of the reactant is zero, thus the kinetic energy of the products will be equal to the Q value of the reaction.

From section (a) the Q value of the reaction is $37.7\text{MeV}$. Thus the kinetic energy of the products is $37.7\text{MeV}$.

Conclusion:

The kinetic energy of the products is $37.7\text{MeV}$.

(c)

To determine

The total momentum shared by the proton and the ${\pi }^{-}$ meson.

Answer to Problem 13P

The total momentum shared by the proton and the ${\pi }^{-}$ meson will be zero.

Explanation of Solution

The ${\Lambda }^0$ particle is at rest. So the momentum of the reactant is zero. As per the conservation of momentum, the total momentum of the products have to be zero in order to conserve the momentum.

Thus the total momentum shared by the proton and the ${\pi }^{-}$ meson will be zero.

Conclusion:

The total momentum shared by the proton and the ${\pi }^{-}$ meson will be zero.

(d)

To determine

Whether the proton and the ${\pi }^{-}$ meson have same kinetic energy.

Answer to Problem 13P

The ${\pi }^{-}$ meson has more kinetic energy than the proton.

Explanation of Solution

The momentum of the proton and ${\pi }^{-}$ meson is same. However, as their masses are different, their kinetic energy will be different. Since the mass of ${\pi }^{-}$ meson is too less than the mass of proton, the ${\pi }^{-}$ meson will have more kinetic energy than the proton.

Write the formula for the kinetic energy of the proton.

    $K_p=\sqrt{m_p^2{c}^4+p^2{c}^2}-m_p{c}^2$                                                                                  (I)

Here, $K_p$ is the kinetic energy of the proton, $m_p$ is the mass of the proton, $c$ is the speed of light in vacuum, and $p$ is the momentum.

Write the formula for the kinetic energy of the ${\pi }^{-}$ meson.

    $K_{\pi }=\sqrt{m_{\pi }^2{c}^4+p^2{c}^2}-m_{\pi }{c}^2$                                                                                    (II)

Here, $K_{\pi }$ is the kinetic energy of the ${\pi }^{-}$ meson, and $m_{\pi }$ is the mass of ${\pi }^{-}$ meson.

The sum of kinetic energy of the proton and ${\pi }^{-}$ meson is equal to the Q value of the reaction.

    $\begin{array}{c}\sqrt{m_p^2{c}^4+p^2{c}^2}-m_p{c}^2+\sqrt{m_{\pi }^2{c}^4+p^2{c}^2}-m_{\pi }{c}^2=Q\\ =m_{\Lambda }{c}^2-m_p{c}^2-m_{\pi }{c}^2\end{array}$

Solve the above equation.

    $\begin{array}{c}\sqrt{m_p^2{c}^4+p^2{c}^2}=m_{\Lambda }{c}^2-\sqrt{m_{\pi }^2{c}^4+p^2{c}^2}\\ m_p^2{c}^4+p^2{c}^2=m_{\Lambda }^2{c}^4-2m_{\Lambda }{c}^2\sqrt{m_{\pi }^2{c}^4+p^2{c}^2}+m_{\pi }^2{c}^4+p^2{c}^2\\ \sqrt{m_{\pi }^2{c}^4+p^2{c}^2}=\frac{m_{\Lambda }^2{c}^4-m_p^2{c}^4+m_{\pi }^2{c}^4}{2m_{\Lambda }{c}^2}\\ pc=\sqrt{{\left(\frac{m_{\Lambda }^2{c}^4-m_p^2{c}^4+m_{\pi }^2{c}^4}{2m_{\Lambda }{c}^2}\right)}^2-m_{\pi }^2{c}^4}\end{array}$                                   (III)

Substitute $1115.6\text{MeV}$ for $m_{\Lambda }{c}^2$, $938.3\text{MeV}$ for $m_p{c}^2$, $139.6\text{MeV}$ for $m_{\pi }{c}^2$ in equation (III).

    $\begin{array}{c}pc=\sqrt{{\left(\frac{{1115.6}^2-{938.3}^2+{139.6}^2}{2(1115.6)}\text{MeV}\right)}^2-{\left(139.6\text{MeV}\right)}^2}\\ =\text{100}\text{.4}\text{MeV}\end{array}$

Substitute $\text{100}\text{.4}\text{MeV}$ for $pc$, $938.3\text{MeV}$ for $m_p{c}^2$ in equation (I).

    $\begin{array}{c}{K}_p=\sqrt{{\left(938.3\text{MeV}\right)}^2+{\left(100.4\text{MeV}\right)}^2}-938.3\text{MeV}\\ =5.35\text{MeV}\end{array}$

Substitute $\text{100}\text{.4}\text{MeV}$ for $pc$, $139.6\text{MeV}$ for $m_{\pi }{c}^2$ in equation (II).

    $\begin{array}{c}{K}_{\pi }=\sqrt{{\left(139.6\text{MeV}\right)}^2+{\left(100.4\text{MeV}\right)}^2}-139.6\text{MeV}\\ =32.3\text{MeV}\end{array}$

Thus the ${\pi }^{-}$ meson has more kinetic energy than the proton.

Conclusion:

The ${\pi }^{-}$ meson has more kinetic energy than the proton.