Chapter 32, Problem 2Q

To determine

If a proton is moving at very high speed, so that its kinetic energy is much greater than it rest energy $\left(m{c}^2\right)$, can it then decay via $P\to n+{\pi }^{+}?$

Answer to Problem 2Q

Solution:

If a proton is moving at very high speed, so that its kinetic energy is much greater than its rest mass energy $\left(m{c}^2\right)$ at higher energy than $m{c}^2$, this decay is possible.

Explanation of Solution

The reaction is given by, $P\to n+{\pi }^{-}$

The energy released is given by

$m_p=1.00727647u=9380272\text{meV}/{\text{c}}^{\text{2}}$

$m_n=1.008664$ u

$=939.565\text{MeV}/{\text{c}}^{\text{2}}$

${{\displaystyle m}}_{{\pi }^{-}}=139.571\text{MeV}/{\text{c}}^{\text{2}}$

$\begin{array}{l}{{\displaystyle E}}_p=m_p{c}^2=938.272\text{MeV}\\ {{\displaystyle E}}_n=mn{c}^2=939.565\text{MeV}\\ {{\displaystyle E}}_{\pi }=m\pi c^2=139.571\text{MeV}\end{array}$

Difference between initial mass energy and final mass energy is the kinetic energy of product.

$\therefore \left({{\displaystyle k}}_n+{{\displaystyle k}}_{{\pi }^{-}}\right)=\left(-938.272+939.565+139.571\right)$

$=+140.864\text{MeV}$

So, at normal energy, this decay is not possible.

Given that energy of proton >> $m{c}^2$

Momentum conservation,

$\left|{{\displaystyle p}}_A\right|=\left|{{\displaystyle p}}_{\pi }\right|=p$

$\Rightarrow \left(\sqrt{{(939.565)}^2}+p^2{c}^2-939.565\right)+\left(\sqrt{{(139.571)}^2+p^2{c}^2}-139.571\right)=140.864$

$\Rightarrow {{\displaystyle p}}_n\pi ={{\displaystyle p}}_p\pi =206.4\text{MeV}$

So, ${{\displaystyle k}}_n=\sqrt{{\left({{\displaystyle m}}_n{c}^2\right)}^2+{\left(206.4\right)}^2}-{{\displaystyle m}}_n{c}^2$

${{\displaystyle k}}_{\pi }=\sqrt{{(139.571)}^2+{(206.4)}^2}-139.571$

$=109.589\text{MeV}.$

Since, ${{\displaystyle k}}_n+{{\displaystyle k}}_{\pi }=131.993\text{MeV}<140\text{MeV}$

So, at higher energy than $m{c}^2$,

This decay is possible.