Chapter 46, Problem 7P

(a)

To determine

Prove that $d\approx \frac{1240}{4\pi m{c}^2}=\frac{98.7}{m{c}^2}$.

Answer to Problem 7P

It is proved that $d\approx \frac{1240}{4\pi m{c}^2}=\frac{98.7}{m{c}^2}$.

Explanation of Solution

Write the uncertainty relation in energy.

    $\Delta E\Delta t=\frac{\hslash }{2}$                                                                                                       (I)

Here, $\Delta E$ is the uncertainty in energy, $\Delta t$ is the uncertainty in time and $\hslash$ is the reduced Plank’s constant.

Write the equation for energy.

    $\Delta E=m{c}^2$                                                                                                       (II)

Here, $m$ is the mass and $c$ is the speed of light.

Substitute equation (II) in (I) and $\frac{h}{2\pi }$ for $\hslash$ and rearrange to find $\Delta t$.

    $\begin{array}{c}m{c}^2\Delta t=\frac{h}{4\pi }\\ \Delta t=\frac{h}{4\pi m{c}^2}\end{array}$                                                                                              (III)

Here, $h$ is the Plank’s constant.

Write the equation for distance travelled by the particle.

    $d\approx c\Delta t$

Here, $d$ is the range.

Substitute equation (III) in the above equation to find $d$.

    $\begin{array}{c}d\approx c\frac{h}{4\pi m{c}^2}\\ =\frac{hc}{4\pi m{c}^2}\end{array}$

Conclusion:

The particle’s rest energy is $m{c}^2$ and its value is $1.602\times {10}^{-19}\text{J/eV}$.

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, and $2.998\times {10}^8\text{m/s}$ for $c$.

    $\begin{array}{c}d\approx \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(2.998\times {10}^8\text{m/s}\right)}{4\pi m{c}^2}\\ =\frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(2.998\times {10}^8\text{m/s}\right)}{4\pi m{c}^2\left(1.602\times {10}^{-19}\text{J/eV}\right)}\\ =\left(\frac{1.240\times {10}^{-6}\text{eV}\cdot \text{m}}{4\pi m{c}^2}\right)\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\\ =\frac{1240\text{eV}\cdot \text{nm}}{4\pi m{c}^2}\end{array}$

Rearrange the above equation.

  $d=\frac{98.7\text{eV}\cdot \text{nm}}{m{c}^2}$

Here, $d$ is in nanometres and $m{c}^2$ is in electron volts.

Thus, it is proved that $d\approx \frac{1240}{4\pi m{c}^2}=\frac{98.7}{m{c}^2}$.

(b)

To determine

The relation between the range and the mass.

Answer to Problem 7P

The range is inversely proportional to the mass of the particle.

Explanation of Solution

Write the relation for the range.

    $d=\frac{98.7\text{eV}\cdot \text{nm}}{m{c}^2}$

Conclusion:

From the above equation, it is clear that the range and mass are inversely proportional.

Thus, the relation between the range and the mass is that the range is inversely proportional to the mass of the particle.

(c)

To determine

The range of the force that might be produced by the virtual exchange of proton.

Answer to Problem 7P

The range of the force that might be produced by the virtual exchange of proton is $\sim {10}^{-16}\text{m}$.

Explanation of Solution

Write the relation for the range.

    $d=\frac{98.7\text{eV}\cdot \text{nm}}{m{c}^2}$

Conclusion:

Substitute $938.3\times {10}^6\text{eV}$ for $m{c}^2$.

    $\begin{array}{c}d=\frac{98.7\text{eV}\cdot \text{nm}}{938.3\times {10}^6\text{eV}}\\ =\left(\frac{98.7\text{nm}}{938.3\times {10}^6}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\\ =1.05\times {10}^{-16}\text{m}\\ \sim {10}^{-16}\text{m}\end{array}$

Thus, the range of the force that might be produced by the virtual exchange of proton is $\sim {10}^{-16}\text{m}$.