Chapter 32, Problem 29P

To determine

Mean life of the measured width of the $\Psi$$\left(3686\right)$ meson which is about $300kev.$

Answer to Problem 29P

Solution:

The mean life time of the meson particle is $2.2\times {10}^{-21}s.$

Explanation of Solution

Given info:

$\Delta E=300kev$

Formula used:

Heisenberg’s uncertainty principle can be expressed as:

$\Delta E=\frac{h}{2\pi \Delta t}$

$\begin{array}{l}\Delta E:\text{- Energy measure}\text{.}\\ h:\text{ Planck's constant}\\ \Delta t:\text{Mean life}\text{.}\end{array}$

Calculation:

Substitute

$\Delta E=300kev,h=6.63\times {10}^{34}J-S$

$\begin{array}{l}300kev=\frac{6.63\times {10}^{-34}}{2\pi \times \Delta t}\\ 300kev\times \frac{1.6\times {10}^{-16}J}{1kev}=\frac{6.63\times {10}^{-34}}{2\pi \times \Delta t}\end{array}$

$\begin{array}{l}4.8\times {10}^{-14}J=\frac{6.63\times {10}^{-34}}{2\pi \times \Delta t}\\ \Delta t=\frac{6.63\times {10}^{-34}J-S}{2\pi \times 4.8\times {10}^{-14}J}\\ =2.2\times {10}^{-21}S.\end{array}$

Therefore, the mean life time of the meson particle is $2.2\times {10}^{-21}s.$