Chapter 19, Problem 47Q

To determine

(a)

The average speed of a Hydrogen atom in the Sun’s atmosphere.

Answer to Problem 47Q

$11.9\text{km}{\text{s}}^{\text{-1}}$.

Explanation of Solution

Given:

Temperature of atmosphere of present day Sun is $T=5800\text{K}$

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

Boltzmann constant $=\text{1}{\text{.38×10}}^{\text{-23}}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}{\text{K}}^{\text{-1}}$

Formula used:

Average speed of an atom is given by the formula,

$\sqrt{\overline{v^2}}=v_{rms}=\sqrt{\frac{3KT}{M}}$

Where

$\sqrt{\overline{v^2}}=v_{rms}$ is root mean square value of speed, M is the mass, K is the Boltzmann’s constant and T is the temperature.

Calculation:

Substitute the values

$\begin{array}{c}{v}_{rms}=\sqrt{\frac{\text{3×1}{\text{.38×10}}^{\text{-23}}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}{\text{K}}^{\text{-1}}\text{×}\text{5800}\text{K}}{\text{1}{\text{.67×10}}^{\text{-27}}\text{kg}}}\\ =11,991.01\text{m}{\text{s}}^{\text{-1}}\\ =11.9\text{km}{\text{s}}^{\text{-1}}\end{array}$

Conclusion:

The average speed of a Hydrogen atom in the Sun’s atmosphere where temperature is $5800K$ is $11.9\text{km}{\text{s}}^{\text{-1}}$.

To determine

(b)

The average speed of a Hydrogen atom in the $1-{\text{M}}_{\odot }$ red giant’s atmosphere.

Answer to Problem 47Q

$9.3\text{km}{\text{s}}^{\text{-1}}$.

Explanation of Solution

Given:

Temperature of atmosphere of $1-{\text{M}}_{\odot },T= 3500\text{K}$

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

Boltzmann constant $=\text{1}{\text{.38×10}}^{\text{-23}}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}{\text{K}}^{\text{-1}}$

Formula used:

Average speed of an atom is given by the formula,

$\sqrt{\overline{v^2}}=v_{rms}=\sqrt{\frac{3KT}{M}}$

Where

$\sqrt{\overline{v^2}}=v_{rms}$ is root mean square value of speed, M is the mass, K is the Boltzmann’s constant and T is the temperature.

Calculation:

$\begin{array}{c}{v}_{rms}=\sqrt{\frac{\text{3×1}{\text{.38×10}}^{\text{-23}}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}{\text{K}}^{\text{-1}}\text{×}\text{3500}\text{K}}{\text{1}{\text{.67×10}}^{\text{-27}}\text{kg}}}\\ \text{=}\text{9,314}\text{.85}\text{m}{\text{s}}^{\text{-1}}\\ \text{=}\text{9}\text{.3}\text{km}{\text{s}}^{\text{-1}}\end{array}$

Conclusion:

The average speed of a Hydrogen atom in the $1-{\text{M}}_{\odot }$ red giant’s atmosphere where temperature is $3500\text{K}$ is $9.3\text{km}{\text{s}}^{\text{-1}}$.

To determine

(c)

Comparison of the two average values of a Hydrogen atom in the atmospheres of present-day Sun and the $1-{\text{M}}_{\odot }$ red giant.

Explanation of Solution

Given:

Escape speed from a red giant = $61.9\text{km}{\text{s}}^{\text{-1}}$

Escape speed from the present-day sun = $619\text{km}{\text{s}}^{\text{-1}}$

According to the calculations, the escape speed of an atom in the atmosphere of a red giant is much lower than the escape speed from the atmosphere of the present-day Sun. However, there is not much difference in the average speeds in the respective atmospheres as they are $9.3\text{km}{\text{s}}^{\text{-1}}$ in the red giant’s atmosphere and $11.9\text{km}{\text{s}}^{\text{-1}}$ at the present-day Sun’s atmosphere. Even though more mass could have escaped to the outer space from a red giant than from a main-sequence star, Hydrogen atoms in both these atmospheres are not having the required speed in order to leave the gravitational field. Hence, both of these stars can retain their respective amounts of Hydrogen atoms.