Chapter 32, Problem 32P

To determine

$\left(a\right)$

What is the energy width of $n^0.$

Answer to Problem 32P

Solution:

The energy width of $n^0$ is $1300ev.$

Explanation of Solution

Given info:

Particle $=n^0$

Concept used:

In a reaction charge, Baryon number, lepton number and strangeness number are conserved.

Baryon number: proton and neutrons both have baryons.

$\begin{array}{l}\text{Lepton number Le}\\ {\overline{V}}_e-1\\ e^{+}-1\\ {\overline{V}}_{\mu }-1\\ {\mu }^{+}-1\\ {\overline{V}}_{\Sigma }-1\\ {\Sigma }^{+}-1\\ V_e\text{ 1}\\ {\text{e}}^{-}\text{ 1}\\ V_{\mu }\text{ 1}\\ {\mu }^{-}1\\ V_{\Sigma }1\\ {\Sigma }^{-}1\end{array}$

Strangeness number:

Strangeness number of quarks up, down and strange.

$\begin{array}{l}\text{ strangeness}\\ \text{U 0}\\ \text{d 0}\\ \text{s }-\text{1}\end{array}$

Calculation:

Substitute

$\begin{array}{c}h=6.63\times {10}^{-34}J-S,\Delta t=5.1\times {10}^{-19}s.\\ \Delta E=\frac{6.63\times {10}^{-34}J-S}{2\pi \times \left(5.1\times {10}^{-19}S\right)}\\ =2.069\times {10}^{-16}J\\ =2.069\times {10}^{-16}J\times \frac{1ev}{1.6\times {10}^{-19}J}\\ =1300ev.\end{array}$

Therefore, the energy width of $n^0=1300ev.$

To determine

$\left(b\right)$

What is the energy width of $p^{+}.$

Answer to Problem 32P

Solution:

The energy width of $p^{+}$ is $150\text{Mev}.$

Explanation of Solution

Given info:

Given particle is $p^{+}$

Concept used:

$B^{+}$ Is the antiparticle of $B^{-}$

$B^0$ Is neutral and must have bottom quark.

${\overline{B}}^0$ Is the antiparticle of $B^0.$

Calculations:

Substitute

$\begin{array}{c}h=6.63\times {10}^{-34}\text{Js},\Delta t=4.4\times {10}^{-24}\text{s}.\\ \Delta E=\frac{6.63\times {10}^{-34}\text{Js}}{2\pi \times \left(4.4\times {10}^{-24}\text{s}\right)}\\ =2.4\times {10}^{-11}\text{J}\\ =2.4\times {10}^{-11}\text{J}\times \frac{1\text{Mev}}{1.6\times {10}^{-13}\text{J}}\\ =150\text{Mev}.\end{array}$