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Found in: Page 1307

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# 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 ${\mathbf{∆}}{\mathbit{p}}$ 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.

The kinetic energy of the nucleon in a nucleus with A = 100 is 30 Me V.

See the step by step solution

## Step 1: Given data

• 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

## Step 2: Understanding the concept of uncertainty principle

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, $r={r}_{0}{A}^{1/3}......\left(3\right)$ $\mathrm{where}{\mathrm{r}}_{0}=1.2\mathrm{fm}$

## Step 3: Calculation of the kinetic energy of the nucleon

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:

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

$K=\frac{{h}^{2}}{2m{r}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(hc\right)}^{2}}{\left(2m{c}^{2}\right){\left({r}_{0}{A}^{1/3}\right)}^{2}}\left(\because \mathrm{From}\mathrm{equation}\left(3\right),\mathrm{r}={\mathrm{r}}_{0}{\mathrm{A}}^{1/3}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\left(1240\mathrm{MeV}.\mathrm{fm}\right)}^{2}}{\left(2\left(931.5\mathrm{MeV}\right)\right){\left(\left(1.2\mathrm{fm}\right)\left({100}^{1/3}\right)\right)}^{2}}\phantom{\rule{0ex}{0ex}}\approx 30\mathrm{MeV}$

Hence, the value of energy is 30 MeV.