Question

From the masses of the weak bosons given in Table 12.1, show that range is of weak part of electroweak force should be about ${\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{\text{\hspace{0.17em}fm}}$.

Short answer

The range is indeed of order${10}^{-3}\text{\hspace{0.17em}fm}$ .

Step-by-step solution

Given data

The force is electroweak.

Concept of the Uncertainty principle

From the Uncertainty principle:

$\mathit{\Delta }\mathit{E}\mathit{\Delta }\mathit{t}\mathbf{\approx }\mathit{\hslash }$

An event in which an amount of energy $\mathit{\Delta }\mathit{E}$ is not conserved is not prohibited as long as the duration of the event does not exceed $\mathit{h}\mathbf{/}\mathit{\Delta }\mathit{E}$.

Let’s assume that a meson particle travels between nucleons at a speed of$\mathit{x}\mathbf{\approx }\mathit{c}$

the temporary energy discrepancy of $\mathit{\Delta }\mathit{E}\mathbf{\approx }\mathit{m}{\mathit{c}}^{\mathbf{2}}$ and that is:

$\mathit{\Delta }\mathit{E}\mathit{\Delta }\mathit{t}\mathbf{\approx }\mathit{\hslash }$ …… (1)

And the time needed for the particle to travel is $\mathit{\Delta }\mathit{t}\mathbf{=}\mathit{x}\mathbf{/}\mathit{c}$.

Equation (1) becomes:

$\mathbf{\left(}\mathbf{m}{\mathbf{c}}^{\mathbf{2}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{x}}{\mathbf{c}}\mathbf{\right)}\mathbf{\approx }\mathit{\hslash }$

Rearrange the above equation for mass $\mathit{m}$and x as:

$\begin{array}{c}\mathbf{x}\mathbf{\approx }\frac{\mathbf{\hslash }}{\mathbf{m}\mathbf{c}}\\ \mathbf{m}\mathbf{\approx }\frac{\mathbf{\hslash }}{\mathbf{x}\mathbf{c}}\end{array}$

Explanation to show electroweak force is  10−3 fm

The mass of the mediation particle of the weak force is about $90\times {10}^3\frac{\text{MeV}}{{\text{c}}^2}$.

Convert units of mass from $\text{MeV}/{\text{c}}^2$ into .

$\begin{array}{l}m=(90\times {10}^3\text{\hspace{0.17em}MeV}/{\text{c}}^2)\left(\frac{1u}{931.5\text{\hspace{0.17em}MeV}/{\text{c}}^2}\right)\\ m=\left(96.6183\text{u}\right)\left(\frac{1.67\times {10}^{-27}\text{\hspace{0.17em}kg}}{\text{u}}\right)\\ m=1.613\times {10}^{-25}\text{kg}\end{array}$

Substitute $1.055\times {10}^{-34}\text{J}\cdot \text{s}$ for$\hslash ,1.613\times {10}^{-25}\text{\hspace{0.17em}kg}$ for$m$ , and $3\times {10}^8\text{\hspace{0.17em}m/s}$for $c$in equation,

$\text{range}\approx \frac{h}{\text{mc}}$

$\begin{array}{l}\text{range}\approx \frac{1.055\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}}{(1.613\times {10}^{-3}\text{\hspace{0.17em}kg})(3\times {10}^8\text{\hspace{0.17em}m/s})}\\ \text{range}\approx 2.197\times {10}^{-18}\text{\hspace{0.17em}m}\end{array}$

Convert units of range from m to as:

$\begin{array}{l}\text{range}\approx (2.197\times {10}^{-18}\text{\hspace{0.17em}m})\left(\frac{1\text{\hspace{0.17em}fm}}{{10}^{-13}\text{\hspace{0.17em}m}}\right)\\ \text{range}\approx 2.197\times {10}^{-3}\text{\hspace{0.17em}fm}\end{array}$

Thus, the range is indeed of order ${10}^{-3}\text{\hspace{0.17em}fm}$.