Chapter 26, Problem 24Q

(a)

To determine

The rest energy of Boson (W) in GeV if the mass of intermediate vector Boson $\left({\text{W}}^{+}\right)$ and the antiparticle of the Boson $\left({\text{W}}^{-}\right)$ is 85.6 times the mass of proton. There exists weak nuclear force in the exchange of Bosons $\left({\text{W}}^{-}\&{\text{W}}^{+}\right)$.

Answer to Problem 24Q

Solution:

80.27 GeV.

Explanation of Solution

Given data:

The mass of intermediate Boson is 85.6 times the mass of a proton.

Formula used:

Einstein proposed in his special theory of relativity that energy is also given as:

$E=m{c}^2$

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

The conversion formula from joule to GeV is:

$1\text{ J}=\frac{1\text{ GeV}}{1.602\times {10}^{-10}}$

Explanation:

Consider the mass of protons to be $1.67\times {10}^{-27}\text{ kg}$ and the speed of light to be $3.0\times {10}^8\text{ m/s}$.

Recall the expression for energy.

$E=m{c}^2$

Substitute $85.6\times \left(1.67\times {10}^{-27}\text{ kg}\right)$ for m and $3.0\times {10}^8\text{ m/s}$ for c. Also, use the conversion formula.

$\begin{array}{c}E=85.6\times \left(1.67\times {10}^{-27}\text{ kg}\right)\times {\left(3.0\times {10}^8\text{ m/s}\right)}^2\\ =1.286\times {10}^{-8}\text{ J}\times \frac{1\text{ GeV}}{1.602\times {10}^{-10}\text{ J}}\\ =80.27\text{ GeV}\end{array}$

Conclusion:

The rest mass energy as given by Einstein’s equation is 80.27 GeV.

(b)

To determine

The rest energy of Boson (W) in GeV if the mass of intermediate vector Boson $\left({\text{W}}^{+}\right)$ and the antiparticle of the Boson $\left({\text{W}}^{-}\right)$ is 85.6 times the mass of proton. There exists weak nuclear force in the exchange of Bosons $\left({\text{W}}^{-}\&{\text{W}}^{+}\right)$.

Answer to Problem 24Q

Solution:

$80.27\times {10}^{13}\text{ K}$.

Explanation of Solution

Given data:

The mass of intermediate Boson is 85.6 times the mass of a proton.

Formula used:

Einstein proposed in his special theory of relativity that energy is also given as:

$E=m{c}^2$

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

The energy and temperature relation is:

$\begin{array}{l}E=kT\\ T=\frac{E}{k}\end{array}$

Here, k is the Boltzmann constant and T is the temperature.

Explanation:

If the rest mass of Boson is taken into consideration, the threshold temperature of the Bosons can be calculated.

Consider the Boltzmann constant to be ${10}^{-13}\text{ GeV/K}$.

Recall the expression for energy and temperature.

$T=\frac{E}{k}$

Substitute ${10}^{-13}\text{ GeV/K}$ for k and recall the value of energy from sub-part (a), that is, 80.27 GeV for T,

$\begin{array}{c}T=\frac{E}{k}\\ =\frac{80.27\text{ GeV}}{{10}^{-13}\text{ GeV/K}}\\ =80.27\times {10}^{13}\text{ K}\end{array}$

Conclusion:

The threshold temperature obtained is $80.27\times {10}^{13}\text{ K}$ for mass energy of 80.27 GeV.

(c)

To determine

The time after the Big Bang when Boson particles and antiparticles disappeared with the help of the below figure.

Answer to Problem 24Q

Solution:

${\text{10}}^{-12}\text{ s}$.

Explanation of Solution

Introduction:

When the Big Bang occurred, all the forces bound together. As the temperature started to decrease with time, the forces got separated because the average temperature of the particles decreased.

Explanation:

From sub-part (a), the energy of the particle is 80.27 GeV and, from sub-part (b), the temperature is $80.27\times {10}^{13}\text{ K}$.

So, from the figure, it can be inferred that the time for which the average energy of the particle is 80.27 GeV and the temperature is $80.27\times {10}^{13}\text{ K}$ is more than ${\text{10}}^{-12}\text{ s}$.

Conclusion:

The time for which the temperature obtained is $80.27\times {10}^{13}\text{ K}$ and the rest mass energy is 80.27 GeV is more than ${\text{10}}^{-12}\text{ s}$.