Chapter 16, Problem 9Q

To determine

(a)

The duration for which the Sun should shine to release energy having an amount equal to that produced by the complete mass-to-energy conversion of a carbon atom.

Answer to Problem 9Q

The duration for which the Sun should shine is $4.6\times {10}^{-36}\text{s}$.

Explanation of Solution

Given:

The luminosity of the Sun is $L_{\odot }=3.90\times {10}^{26}\text{W}$.

The mass of the carbon atom is $m=2\times {10}^{-26}\text{kg}$.

Formula Used:

The duration for which the Sun should shine is given by

$t=\frac{E}{L_{\odot }}$

Here, $E$ is the energy released as a result of mass-energy equivalence.

Einstein’s mass-energy relation is given by

$E=m{c}^2$

Calculations:

The energy released as a result of mass-energy equivalence is calculated as

$\begin{array}{c}E=m{c}^2\\ =\left(2\times {10}^{-26}\text{kg}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =1.8\times {10}^{-9}\text{J}\end{array}$

The duration for which the Sun should shine is calculated as

$\begin{array}{c}t=\frac{E}{L_{\odot }}\\ =\frac{1.8\times {10}^{-9}\text{J}}{3.90\times {10}^{26}\text{W}}\\ =4.6\times {10}^{-36}\text{s}\end{array}$

Conclusion:

The duration for which the Sun should shine to release energy having an amount equal to that produced by the complete mass-to-energy conversion of a carbon atom is $4.6\times {10}^{-36}\text{s}$.

To determine

(b)

The duration for which the Sun should shine to release energy having an amount equal to that produced by the complete mass-to-energy conversion of $1\text{kg}$ substance.

Answer to Problem 9Q

The duration for which the Sun should shine is $2.3\times {10}^{-10}\text{s}$.

Explanation of Solution

Given:

The luminosity of the Sun is $L_{\odot }=3.90\times {10}^{26}\text{W}$.

The mass of the substance is $m=1\text{kg}$.

Formula Used:

The duration for which the Sun should shine is given by

$t=\frac{E}{L_{\odot }}$

Here, $E$ is the energy released as a result of mass-energy equivalence.

Einstein’s mass-energy relation is given by

$E=m{c}^2$

Calculations:

The energy released as a result of mass-energy equivalence is calculated as

$\begin{array}{c}E=m{c}^2\\ =\left(1\text{kg}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =9\times {10}^{16}\text{J}\end{array}$

The duration for which the Sun should shine is calculated as

$\begin{array}{c}t=\frac{E}{L_{\odot }}\\ =\frac{9\times {10}^{16}\text{J}}{3.90\times {10}^{26}\text{W}}\\ =2.3\times {10}^{-10}\text{s}\end{array}$

Conclusion:

The duration for which the Sun should shine to release energy having an amount equal to that produced by the complete mass-to-energy conversion of $1\text{kg}$ substance is $2.3\times {10}^{-10}\text{s}$.

To determine

(c)

The duration for which the Sun should shine to release energy having an amount equal to that produced by the complete mass-to-energy conversion of Earth.

Answer to Problem 9Q

The duration for which the Sun should shine is $1.38\times {10}^{15}\text{s}$.

Explanation of Solution

Given:

The luminosity of the Sun is $L_{\odot }=3.90\times {10}^{26}\text{W}$.

The mass of the Earth is $m=6\times {10}^{24}\text{kg}$.

Formula Used:

The duration for which the Sun should shine is given by

$t=\frac{E}{L_{\odot }}$

Here, $E$ is the energy released as a result of mass-energy equivalence.

Einstein’s mass-energy relation is given by

$E=m{c}^2$

Calculations:

The energy released as a result of mass-energy equivalence is calculated as

$\begin{array}{c}E=m{c}^2\\ =\left(6\times {10}^{24}\text{kg}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =5.4\times {10}^{41}\text{J}\end{array}$

The duration for which the Sun should shine is calculated as

$\begin{array}{c}t=\frac{E}{L_{\odot }}\\ =\frac{5.4\times {10}^{41}\text{J}}{3.90\times {10}^{26}\text{W}}\\ =1.38\times {10}^{15}\text{s}\end{array}$

Conclusion:

The duration for which the Sun should shine to release energy having an amount equal to that produced by the complete mass-to-energy conversion of Earth is $1.38\times {10}^{15}\text{s}$.