Because a nucleon is confined to a nucleus, we can take the uncertainty in its position to be approximately the nuclear radius r. Use the uncertainty $\mathbf{∆}\mathit{p}$ principle to determine the uncertainty in the linear momentum of the nucleon. Using the approximation
$\mathbf{p}\mathbf{\approx }\mathbf{∆}\mathbf{p}$and the fact that the nucleon is non-relativistic, calculate the kinetic energy of the nucleon in a nucleus with A = 100.

• Uncertainty in position is approximately the nuclear radius,$∆\mathrm{x}=\mathrm{r}$
• Approximation of momemtum,$\mathrm{p}\approx ∆\mathrm{p}$
• The nucleon is non-relativistic.
• Mass or nucleon number, A =100

The uncertainty principle states that the position and the velocity of an object cannot be measured simultaneously. Again, we are provided that the nucleon is non-relativistic, thus the energy can be found using the certainty condition of momentum and position considering the confinement of the nucleon with the nucleus.

Formulae:

The uncertainty relation of momentum-position,$∆\mathrm{p}=\frac{\mathrm{h}}{∆\mathrm{x}}......\left(1\right)$

The kinetic energy of a body in motion,$\mathrm{K}=\frac{{\mathrm{p}}^2}{2\mathrm{m}}...........\left(2\right)$

The radius of an atom in a nucleus, $r=r_0{A}^{1/3}......\left(3\right)$ $\mathrm{where}{\mathrm{r}}_0=1.2\mathrm{fm}$

Substituting the value of value of momentum from equation (1) in equation (2), the kinetic energy of the nucleon can be calculated using the given data as follows:

$\left(\mathrm{for}\mathrm{p}\approx ∆\mathrm{p},∆\mathrm{x}=\mathrm{r},\mathrm{hc}=1240\mathrm{MeV}.\mathrm{fm},{\mathrm{mc}}^2=931.5\mathrm{MeV}\right)$

$$\begin{gathered} K=\frac{h^2}{2m{r}^2} \\ =\frac{{\left(hc\right)}^2}{\left(2m{c}^2\right){\left(r_0{A}^{1/3}\right)}^2}\left(\because \mathrm{From}\mathrm{equation}\left(3\right),\mathrm{r}={\mathrm{r}}_0{\mathrm{A}}^{1/3}\right) \\ =\frac{{\left(1240\mathrm{MeV}.\mathrm{fm}\right)}^2}{\left(2\left(931.5\mathrm{MeV}\right)\right){\left(\left(1.2\mathrm{fm}\right)\left({100}^{1/3}\right)\right)}^2} \\ \approx 30\mathrm{MeV} \end{gathered}$$

Hence, the value of energy is 30 MeV.