Chapter 19, Problem 47Q

(a)

To determine

The average speed of a hydrogen atom having mass =$1.67\times {10}^{-27}\text{kg}$ in the atmosphere of the present-day Sun, given the temperature of the Sun is 5800 K.

Answer to Problem 47Q

Solution: $1.20\times {10}^4\text{m/s}$

Explanation of Solution

Given data:

For present-day Sun, $T=5800\text{ K}\text{.}$

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

Formula used:

Write the expression for the average speed for a gas atom or molecule.

$v=\sqrt{\frac{3kT}{m}}$

Here,

$v$= the average speed of a gas atom or molecule in $\text{m/s}$

$k=1.38\times {10}^{-23}\text{J/K}$

$T$= the temperature of a gas in kelvins

$m$= the mass of atom or molecule in kilograms

Explanation:

Recall the equation for the average speed.

$v=\sqrt{\frac{3kT}{m}}$

Substitute $5800\text{ K for }T,1.67\times {10}^{-27}\text{kg for }m$

$\begin{array}{c}v=\sqrt{\frac{3\times 1.38\times {10}^{-23}\text{ J/K}\times 5800\text{ K}}{1.67\times {10}^{-27}\text{kg}}}\\ =1.20\times {10}^4\text{ m/s}\text{.}\end{array}$

Conclusion:

Therefore, the average speed of the hydrogen atom is $1.20\times {10}^4\text{m/s}\text{.}$

(b)

To determine

The average speed of a hydrogen atom having mass =$1.67\times {10}^{-27}\text{kg}$ in the atmosphere of ${\text{1-M}}_{\text{Θ}}$ red giant having a temperature of 3500 K.

Answer to Problem 47Q

Solution: $9.31\times {10}^3\text{m/s}$

Explanation of Solution

Given data:

For ${\text{1-M}}_{\text{Θ}}$ red giant $T=3500\text{ K}\text{.}$

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

Formula used:

Write the expression for the average speed for a gas atom or molecule.

$v=\sqrt{\frac{3kT}{m}}$

Here,

$v$= the average speed of a gas atom or molecule in $\text{m/s}$

$k=1.38\times {10}^{-23}\text{J/K}$

$T$= the temperature of a gas in kelvins

$m$= the mass of atom or molecule in kilograms

Explanation:

Recall the equation for the average speed.

$v=\sqrt{\frac{3kT}{m}}$

Substitute $3500\text{ K for }T,1.67\times {10}^{-27}\text{kg for }m$

$\begin{array}{c}v=\sqrt{\frac{3\times 1.38\times {10}^{-23}\text{ J/K}\times 3500\text{ K}}{1.67\times {10}^{-27}\text{kg}}}\\ =9.31\times {10}^3\text{ m/s}\text{.}\end{array}$

Conclusion:

Therefore, the average speed of the hydrogen atom in the atmosphere of the red giant is $9.31\times {10}^3\text{m/s}\text{.}$

(c)

To determine

The extent to which the present day Sun and a ${\text{1-M}}_{\text{Θ}}$ red giant can retain hydrogen in their atmosphere based on the escape speed and the average speed for both conditions.

Answer to Problem 47Q

Solution: In both cases, the value of the escape speed is higher than the average speed of the atoms but is close in the case of a red giant. Therefore, one can say that for speed above the average speed, hydrogen will leave the red giant.

Explanation of Solution

Introduction:

The escape velocity is the minimum velocity required to escape the gravitational attraction of a planet or star.

Explanation:

Calculate the escape velocity for both cases.

For present-day Sun, $M=\text{1}{\text{.99×10}}^{\text{30}}\text{kg, }R=696,000\text{ km}=6.96\times {10}^8\text{m}\text{.}$

Write the expression for the escape speed.

$v_{esc}=\sqrt{\frac{2GM}{R}}$

Here, $M\text{ is mass of Sun, }R\text{ is radius of Sun and G is a constant having value }6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Recall the equation for the escape speed.

$v_{esc}=\sqrt{\frac{2GM}{R}}$

Substitute $1.99\times {10}^{30}\text{kg for }M\text{, 6}\text{.96}\times {\text{10}}^{8}6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\text{ for G}\text{.}$

$\begin{array}{c}{v}_{esc}=\sqrt{\frac{2\times 6.67\times {10}^{-11}\text{ N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}\times 1.99\times {10}^{30}\text{ kg}}{6.96\times {10}^8\text{m}}}\\ =6.29\times {10}^5\text{ m/s}\text{.}\end{array}$

Calculate the escape speed for the red giant;

Recall the equation for the escape speed.

$v_{esc}=\sqrt{\frac{2GM}{R}}$

Substitute $1.99\times {10}^{30}\text{kg for }M\text{, 6}\text{.96}\times {\text{10}}^8\text{m for }R\text{ and }6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\text{ for G}\text{.}$

$\begin{array}{c}{v}_{esc}=\sqrt{\frac{2\times 6.67\times {10}^{-11}\text{ N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}\times 1.99\times {10}^{30}\text{ kg}}{6.96\times {10}^{10}\text{m}}}\\ =6.29\times {10}^4\text{ m/s}\text{.}\end{array}$

The escape velocity for present-day Sun is $6.18\times {10}^5\text{ m/s}$ and that of the red giant phase of the Sun is $6.18\times {10}^4\text{ m/s}\text{.}$ As compared to the average speed of atoms in Sun calculated in (a) is $1.20\times {10}^4\text{m/s}$ and that in case of red giant calculated in (b) is $9.31\times {10}^3\text{m/s}$, we can clearly see that the average speed in both cases is less than the escape speed. Hence both will retain the hydrogen in their cores respectively. But the value of average speed is close to the escape speed in the case of the red giant, hence, it might lose hydrogen more easily as compared to the Sun.

Conclusion:

Hence, both the Sun and the red giant will retain hydrogen in their cores but the red giant is more likely to lose that hydrogen as compared to the Sun.