91影视

The range of the force is${10}^{鈭�20}\text{鈥尘}$ .

Uncertainty principle:

$\mathit{螖}\mathit{E}\mathit{螖}\mathit{t}\mathbf{鈮�}\mathit{鈩�}$

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 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︹赌�(1)

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

An event in which an amount of energy$\mathit{螖}\mathit{E}$ is not conserved is not prohibited as long as the duration

The energy of the particle is given as,

$螖E鈮�\frac{\mathrm{鈩�}c}{x}$

Substitute $1.055脳{10}^{鈭�34}\text{鈥�}J鈰�s$ for$\mathrm{鈩�}$, and$3脳{10}^8\text{鈥�}m/s$ for $c$ and ${10}^{鈭�20}\text{鈥�}m$ for $x$ in the above equation as:

$\begin{array}{l}螖E鈮�\left(\frac{1.055脳{10}^{鈭�34}\text{鈥塉}鈰�\text{s}}{{10}^{鈭�20}\text{鈥尘}}\right)(3脳{10}^8\text{鈥尘/s})\\ 螖E=3.165脳{10}^{鈭�6}\text{鈥�}\text{J}\left(\frac{1.0\text{鈥塭痴}}{1.6脳{10}^{鈭�19}\text{鈥塉}}\right)\\ 螖E=1.97脳{10}^{13}\left(\frac{1\text{鈥涂别痴}}{{10}^{12}\text{鈥塭痴}}\right)\\ 螖E=20\text{鈥涂别痴}\end{array}$

Therefore, we need $20\text{鈥涂别痴}$ kinetic energy to produce this particle.