91影视

Given that the radius of the atomic nucleus is of the order of $5脳{10}^{-15}m$. We can take the uncertainty in position of the proton inside the nucleus, $鈭�r=5脳{10}^{-15}m$.

From the Heisenberg鈥檚 Uncertainty Principal for momentum and position,

$\mathbf{鈭�}{\mathit{p}}_{\mathbf{r}}\mathbf{鈭�}\mathit{r}\mathbf{鈮�}\frac{\mathbf{h}}{\mathbf{2}}$.

The uncertainty in momentum will be

$$\begin{gathered} 鈭�p_r鈮�\frac{\sout{h}}{2鈭�r} \\ 鈭�p_r鈮�\frac{1.055脳{10}^{-34}}{2脳5脳{10}^{-15}} \\ 鈭�p_r鈮�1.05脳{10}^{-20}kg.m/s \end{gathered}$$

So, the uncertainty in the momentum is $鈭�p_r鈮�1.05脳{10}^{-20}kg.m/s$.

Taking the minimum uncertainty in the momentum of proton as an estimated momentum, $p=1.05脳{10}^{-20}kg路m/s$.

Now, solve for $E$.

$$\begin{gathered} E^2={\left[\left(1.05脳{10}^{-20}\right)\left(3脳{10}^8\right)\right]}^2+{\left[\left(1.67脳{10}^{-27}\right)\left(3脳{10}^8\right)\right]}^2 \\ {E}^2=2.26脳{10}^{-20} \\ E=1.5脳{10}^{-10}J \end{gathered}$$

Using the energy, we can find the value of $K$.

$$\begin{gathered} K=E-m{c}^2 \\ K=1.5脳{10}^{-10}-\left[\left(1.67脳{10}^{-27}\right){\left(3脳{10}^8\right)}^2\right] \\ K=3.32脳{10}^{-14}J \end{gathered}$$

For a proton to remain bound to the nucleus, the negative potential energy must be at least equal to the kinetic energy of the proton. Thus,

$$\begin{gathered} U=\frac{3.32脳{10}^{-14}}{1.6脳{10}^{-19}} \\ U=2.08脳{10}^{5}eV \\ U=0.208MeV \end{gathered}$$

This value is about${10}^4$ larger than the potential energy of an electron in a hydrogen atom.