Chapter 30.3, Problem 39PP

(a)

To determine

The energy equivalent of the neutron’s mass in a megaelectron volts.

Answer to Problem 39PP

The energy equivalent of the neutron’s mass in a megaelectron volts is $939.57\text{ MeV}$ .

Explanation of Solution

Given:

Mass of neutron, $m_n=1.00866\text{ u}$

Formula used:

Energy, $E=\left(m\times 931.5\right)\text{ MeV}$

  $1u=931.5\text{ MeV}$

Calculation:

The energy of neutron’s mass is:

  $E=m_n{c}^2\text{ (in Joules)}$

  $\begin{array}{l}E=\left(m\times 931.5\right)\text{ MeV}\\ E=\left(1.00866\times 931.5\right)\text{ MeV}\\ E=939.57\text{ MeV}\end{array}$

Conclusion:

The energy equivalent of the neutron’s mass in a megaelectron volts is $939.57\text{ MeV}$ .

(b)

To determine

The smallest total $\text{γ}$ -ray energy of that could result in the production of neutron-antineutron pair.

Answer to Problem 39PP

The total energy required is $1879.14\text{ MeV}$ .

Explanation of Solution

Given:

The energy equivalent of the neutron’s mass in a megaelectron volts is $939.57\text{ MeV}$ .

Formula used:

The smallest total $\text{γ}$ -ray energy of that could result in the production of neutron-antineutron pair would be equal to:

  $E^{\gamma }=2m_n{c}^2$

Here, c is the speed of light.

Calculation:

The required energy is calculated as,

  $\begin{array}{l}{E}^{\gamma }=2m_n{c}^2\\ E^{\gamma }=2E\\ E^{\gamma }=2\left(939.57\right)\text{ MeV}\\ E^{\gamma }=1879.14\text{ MeV}\end{array}$

Conclusion:

The total energy required is $1879.14\text{ MeV}$ .