Chapter 30.3, Problem 37PP

(a)

To determine

To calculate:

Energy equivalent of the proton's mass in joules.

Answer to Problem 37PP

Energy equivalent of the proton's mass in joules is $1.503\times {10}^{-10}\text{J}$ .

Explanation of Solution

Given:

  $m=1.67\times {10}^{-27}\text{kg}$

Formula used:

  $E=m\times c^2$

Where,

  $E:$ Energy

  $m:$ Mass

  $c:$ Speed of light

Calculation:

Energy equivalent of the proton's mass is calculated by formula,

  $E=m\times c^2$

Substituting the values in above formula,

  $\begin{array}{l}E=\left(1.67\times {10}^{-27}\text{kg}\right)\times {\left(3\times {10}^8\text{m/s}\right)}^2\\ =\left(1.67\times {10}^{-27}\text{kg}\right)\times \left(9\times {10}^{16}\text{m/s}\right)\\ =1.503\times {10}^{-10}\text{J}\end{array}$

Conclusion:

Hence, energy equivalent of the proton's mass in joules is $1.503\times {10}^{-10}\text{J}$ .

(b)

To determine

To calculate: Conversion of value of energy in electron volts.

Answer to Problem 37PP

The value of energy in electron volts is $939\text{MeV}$ .

Explanation of Solution

Given:

  $E=1.503\times {10}^{-10}\text{J}$

Formula used:

For conversion,

  $1J=6.242\times {10}^{12}\text{MeV}$

Calculation:

The value of energy in electron volts can be given as,

  $E=\left(1.503\times {10}^{-10}\text{J}\right)\times \left(6.242\times {10}^{12}\text{MeV}\right)$

As,

  $\begin{array}{l}1J=6.242\times {10}^{12}\text{MeV}\\ E=939\text{MeV}\end{array}$ ,

Conclusion:

Hence, the value of energy in electron volts is $939\text{MeV}$ .

(c)

To determine

To calculate:

The smallest total $\lambda -ray$ energy that could result in a proton-antiproton pair.

Answer to Problem 37PP

The smallest total $\lambda -ray$ energy that could result in a proton-antiproton pair is $1878\text{MeV}$

Explanation of Solution

Given:

  $E=939\text{MeV}$

Formula used:

  $E_{\lambda }=2\times E$

Where,

  $E_{\lambda }:$

  $\lambda -$ ray energy

  $E:$ Equivalent energy

Calculation:

The smallest total $\lambda -ray$ energy that could result in a proton-antiproton pair can be calculated from,

  $E_{\lambda }=2\times E$

Substituting the value in above formula,

  $\begin{array}{l}{E}_{\lambda }=2\times \left(939\text{MeV}\right)\\ =1878\text{MeV}\end{array}$

Conclusion:

Hence, the smallest total $\lambda -ray$ energy that could result in a proton-antiproton pair is $1878\text{MeV}$ .