Question

From the experimental evidence that the force between nucleons has a range of about 1 fm. Obtaina rough value (in MeV/c²) for the mass of the particle exchanged to convey this force, the pion.

Short answer

The mass of the mediation particle is$197{\text{\hspace{0.17em}MeV/c}}^{\text{2}}$ .

Step-by-step solution

Given data

The force between nucleons has a range of about $1\text{\hspace{0.17em}fm}$.

Concept of the Uncertainty principle 

From the Uncertainty principle:

$\Delta E\Delta t\approx \hslash$

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$\mathit{\hslash }\mathbf{/}\mathit{\Delta }\mathit{E}$ .

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 and x as:

Calculation of the mass of the mediation particle

The range of particles is$1\text{\hspace{0.17em}fm}$.

Convert units of distance from to :

$\begin{array}{l}x=\left(1\text{\hspace{0.17em}fm}\right)\left(\frac{1\text{\hspace{0.17em}m}}{{10}^{-15}\text{\hspace{0.17em}m}}\right)\\ x=1\times {10}^{-15}\text{\hspace{0.17em}m}\end{array}$

Substitute$1.055\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}$ for $\hslash ,1\times {10}^{-35}\text{m}$ for $x$and $3\times {10}^8\text{\hspace{0.17em}m/s}$ for $c$ in equation:.

$m\approx \frac{\hslash }{xc}$

$\begin{array}{l}m\approx \frac{1.055\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}}{(1\times {10}^{-15}\text{\hspace{0.17em}m})(3\times {10}^8\text{\hspace{0.17em}m}/\text{s})}\\ m\approx 3.51\times {10}^{-28}\text{kg}\end{array}$

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

$\begin{array}{l}m\approx (3.51\times {10}^{-28}\text{\hspace{0.17em}kg})\left(\frac{1\text{u}}{1.67\times {10}^{-27}\text{kg}}\right)\\ m\approx \left(0.211\text{u}\right)\left(\frac{931.5\text{\hspace{0.17em}MeV}/{\text{c}}^2}{1\text{\hspace{0.17em}u}}\right)\\ m\approx 196.54{\text{\hspace{0.17em}MeV/c}}^{\text{2}}\\ m\approx 197{\text{\hspace{0.17em}MeV/c}}^{\text{2}}\end{array}$

Therefore, the mass of the mediation particle (pion) is$197{\text{\hspace{0.17em}MeV/c}}^{\text{2}}$