91影视

The force between nucleons has a range of about $1\text{鈥塮尘}$.

From the Uncertainty principle:

$螖E螖t鈮�\mathrm{鈩�}$

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

Let鈥檚 assume that a meson particle travels between nucleons at a speed of$\mathit{x}\mathbf{鈮�}\mathit{c}$

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

$\mathit{螖}\mathit{E}\mathit{螖}\mathit{t}\mathbf{鈮�}\mathit{鈩�}$ 鈥︹�� (1)

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

Equation (1) becomes:

$\mathbf{\left(}\mathbf{m}{\mathbf{c}}^{\mathbf{2}}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{x}}{\mathbf{c}}\mathbf{\right)}\mathbf{鈮�}\mathit{鈩�}$

Rearrange the above equation for mass and x as:

The range of particles is$1\text{鈥塮尘}$.

Convert units of distance from to :

$\begin{array}{l}x=\left(1\text{鈥塮尘}\right)\left(\frac{1\text{鈥尘}}{{10}^{鈭�15}\text{鈥尘}}\right)\\ x=1脳{10}^{鈭�15}\text{鈥尘}\end{array}$

Substitute$1.055脳{10}^{鈭�34}\text{鈥塉}鈰�\text{s}$ for $\mathrm{鈩�},\text{鈥�}1脳{10}^{鈭�35}\text{鈥�}\text{m}$ for $x$and $3脳{10}^8\text{鈥尘/s}$ for $c$ in equation:.

$m鈮�\frac{\mathrm{鈩�}}{xc}$

$\begin{array}{l}m鈮�\frac{1.055脳{10}^{鈭�34}\text{鈥塉}鈰�\text{s}}{(1脳{10}^{鈭�15}\text{鈥尘})(3脳{10}^8\text{鈥尘}/\text{s})}\\ m鈮�3.51脳{10}^{鈭�28}\text{鈥�}\text{kg}\end{array}$

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

$\begin{array}{l}m鈮�(3.51脳{10}^{鈭�28}\text{鈥塳驳})\left(\frac{1\text{u}}{1.67脳{10}^{鈭�27}\text{鈥�}\text{kg}}\right)\\ m鈮�\left(0.211\text{u}\right)\left(\frac{931.5\text{鈥塎别痴}/{\text{c}}^2}{1\text{鈥塽}}\right)\\ m鈮�196.54{\text{鈥塎别痴/肠}}^{\text{2}}\\ m鈮�197{\text{鈥塎别痴/肠}}^{\text{2}}\end{array}$

Therefore, the mass of the mediation particle (pion) is$197{\text{鈥塎别痴/肠}}^{\text{2}}$