Chapter 32, Problem 24P

To determine

In the rare decay ${\pi }^{+}\to e^{+}+Ve,$ what is the kinetic energy of the positron?

Answer to Problem 24P

Solution:

The kinetic energy of the positron is $69.3Mev.$

Explanation of Solution

Given info:

${\pi }^{+}\to e^{+}+Ve$

Concept used:

The energy released during reaction can be calculated as $Q=m{c}^2$

Calculation:

The energy released during the reaction $Q=\left[m_{{\pi }^{+}}-m_{e^{+}}-m_{ve}\right]C^2$

Substitute

$\begin{array}{l}{M}_{A^{+}}=0.1498658u\\ m_{e^{-}}=0.000548577U.\end{array}$

$\begin{array}{c}Q=\left[0.1498658-0.000548577\right]C^2\\ =\left[0.149317223\right]C^2\\ =0.149317223\times 931.5\\ =139.0889Mev.\end{array}$

Therefore, the energy released in decay $=139.0889Mev.$

The energy released in any nuclear reaction is equal to the kinetic energy of the final preoduct particles.

$\begin{array}{l}Q=K.E_e+K.E_v\\ K.E_v=Q-K.E_e\end{array}$

Here,

$\begin{array}{l}K.E_v=\text{Kinetic energy of neutrino}\text{.}\\ K.E_e=\text{Kinetic energy of positiron}\text{.}\end{array}$

The expression for the energy of the neutrion is

$\begin{array}{l}{E}_v=\sqrt{{{\displaystyle p}}_V^2{C}^2+{{\displaystyle m}}_v^2{C}^4}\\ {{\displaystyle E}}_v^2={{\displaystyle p}}_V^2{C}^2+{{\displaystyle m}}_v^2{C}^4\end{array}$

${{\displaystyle p}}_V^2{C}^2={{\displaystyle E}}_v^2-{{\displaystyle m}}_v^2{C}^4$

Now, apply conservation of linear momentum.

The momentum of positron is equal to that of neutrino in magnitude but opposite in direction, since the parent particle is initially rest.

$\begin{array}{l}{\left|P_e\right|}^2={\left|P_v\right|}^2\\ {\left|P_e\right|}^2{C}^2={\left|P_v\right|}^2{C}^2\\ {{\displaystyle E}}_e^2-{{\displaystyle m}}_e^2{C}^4={{\displaystyle E}}_v^2-{{\displaystyle m}}_e^2{C}^2\end{array}$

Substitute $K.E_e+m_e{C}^2$ for $E_e$ and $K.E_v+M_v{C}^2$ for $E_v$

${\left(K.E_e+m_e{C}^2\right)}^2-{{\displaystyle m}}_e^2{C}^4={\left(K.E_v+m_v{C}^2\right)}^2-{{\displaystyle m}}_v^2{C}^4$

${\left(K.E_e+m_e{C}^2\right)}^2-{{\displaystyle m}}_e^2{C}^4={\left(K.E_v\right)}^2-0$

$K.{{\displaystyle E}}_e^2+2K.E_e{m}_e{C}^2+{{\displaystyle m}}_e^2{C}^4-{{\displaystyle m}}_e^2{C}^4=K.{{\displaystyle E}}_v^2$

$K.{{\displaystyle E}}_e^2+2\left(K.E_e\right)m_e{C}^2={\left(K.E_e\right)}^2$

Substitute $Q-k.E_e$ for $K.E_v$

$K.{{\displaystyle E}}_e^2+2\left(K.E_e\right)m_e{C}^2={\left(Q-K.E_e\right)}^2$

$K.{{\displaystyle E}}_e^2+2K.E_e{m}_e{C}^2+=Q^2+K.{{\displaystyle E}}_e^2-2K{E}_{e}Q$

$k.E_e=\frac{Q^2}{2m_e{C}^2+2Q}$

Substitute $Q=139.089mev,$$m_e{C}^2=0.511Mev.$

$\begin{array}{l}k.E_e=\frac{{\left(139.089Mev\right)}^2}{2\left(0.511Mev\right)+2\left(139.89Mev\right)}\\ =69.3Mev.\end{array}$

Therefore, the kinetic energy of the positron is $69.3Mev.$