Chapter 16, Problem 11Q

To determine

(a)

The mass of hydrogen consumed by Sun over the past $4.56\text{billion years}$.

Answer to Problem 11Q

The mass of hydrogen consumed by Sun over the past $4.56\text{billion years}$ is $8.7\times {10}^{28}\text{kg}$.

Explanation of Solution

Given:

Time period in which hydrogen is consumed is $4.56\text{billion years}$.

Concept used:

Total amount of energy given off by the Sun over its lifetime to produce the Sun’s Luminosity of $3.9\times {10}^{26}\text{J/s}$ can be calculated by the formula given below.

Write the expression for the relation between energy and luminosity.

$L=\frac{E}{t}$

Here, $L$ is the luminosity of the Sun, $E$ is the energy given off by the Sun over its lifetime to produce the Sun’s Luminosity and $t$ is the time period.

Rearrange above equation for Energy.

$E=L\cdot t$ ....... (1)

Convert $1\text{billion}\text{years}$ into seconds:

$\begin{array}{c}\text{Conversion}\text{factor}\to 1\text{billion}\text{years}\left(\frac{{10}^9}{1\text{billion}}\right)\left(\frac{3.15\times {10}^7\mathrm{s}}{1\text{year}}\right)\\ \to 3.15\times {10}^{16}\mathrm{s}\end{array}$

Six ${}^1\text{H}$ nuclei produces two ${}^3\text{He}$ nuclei, which combines to form one ${}^4\text{He}$ nucleus and two of the original ${}^1\text{H}$ nuclei return back to its original state, the three steps of proton-proton chain can be summarized as:

$4{}^1\text{H}{\to }^4\text{He}+\text{energy}$

Energy lost when four hydrogen combine to form one helium atom is $4.3\times {10}^{-12}\text{J}$.

The mass of hydrogen consumed by Sun is the product of energy given off by the Sun over its lifetime and the fraction of energy lost by the four hydrogen atoms.

Mass of one hydrogen atom is $1.67\times {10}^{-27}\text{kg}$.

Write the expression for mass of hydrogen consumed by Sun.

$m_{\text{hydrogen}}=E\left(\frac{4m_H}{E_{\text{lost}}}\right)$ ....... (2)

Here, $m_{\text{hydrogen}}$ is the mass of hydrogen consumed by Sun, $m_H$ is the mass of one hydrogen atom and $E_{\text{lost}}$ is the amount of energy lost when four hydrogen atoms convert into one helium atom.

Calculation:

Substitute $4.56\text{billion years}$ for $t$ and $3.9\times {10}^{26}\text{watts}$ for $L$ in equation (1).

$\begin{array}{c}E=\left(3.9\times {10}^{26}\text{watts}\right)\cdot \left(4.56\text{billion years}\left(\frac{3.15\times {10}^{16}\mathrm{s}}{1\text{billion}\text{years}}\right)\right)\\ =5.6\times {10}^{43}\text{J}\end{array}$

Substitute $5.6\times {10}^{43}\text{J}$ for $E$, $1.67\times {10}^{-27}\text{kg}$ for $m_H$ and $4.3\times {10}^{-12}\text{J}$ for $E_{\text{lost}}$ in equation (2).

$\begin{array}{c}{m}_{\text{hydrogen}}=\left(5.6\times {10}^{43}\text{J}\right)\left(\frac{4\left(1.67\times {10}^{-27}\text{kg}\right)}{4.3\times {10}^{-12}\text{J}}\right)\\ =8.69\times {10}^{28}\text{kg}\\ =8.7\times {10}^{28}\text{kg}\end{array}$

Conclusion:

Thus,the mass of hydrogen consumed by Sun over the past $4.56\text{billion years}$ is $8.7\times {10}^{28}\text{kg}$.

To determine

(b)

Explain whether answer in part (a) is an overestimate or an underestimate.

Answer to Problem 11Q

The answer in part (a) is an overestimate.

Explanation of Solution

The luminosity of the Sun when it was first formed was only $70\%$ of the present value.

In above part it is considered that luminosity of Sun remained constant over the time period of $4.56\text{billion years}$ which is $3.9\times {10}^{26}\text{J/s}$.

So the amount of hydrogen consumed when Luminosity was only 70% of the present value should be less than the amount consumed now.

Therefore, the value of hydrogen calculated in above part is the overestimate.

Conclusion:

Thus, the answer in part (a) is an overestimate.