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Hydrogen on the Sun. The surface of the sun has a temperature of about 5800 \(\mathrm{K}\) and consists largely of hydrogen atoms. (a) Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is \(1.67 \times 10^{-27}\) kg. (b) The escape speed for a particle to leave the gravitational influence of the sun is given by \((2 G M / R)^{1 / 2},\) where \(M\) is the sun's mass, \(R\) its radius, and \(G\) the gravitational constant \((\) see Example 13.5 of Section 13.3\() .\) Use the data in Appendix \(F\) to calculate this escape speed. (c) Can appreciable quantities of hydrogen escape from the sun? Can any hydrogen escape? Explain.

Short Answer

Expert verified
The RMS speed of a hydrogen atom is about 11800 m/s.

Step by step solution

01

Calculate RMS Speed of Hydrogen

To find the root mean square speed, we use the formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( k \) is the Boltzmann constant \(1.38 \times 10^{-23} \ \text{J/K}\), \( T \) is the temperature in Kelvin, and \( m \) is the mass of a hydrogen atom. Substituting the values:\[ v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 5800}{1.67 \times 10^{-27}}} \]Calculating this gives:\[ v_{rms} \approx \sqrt{\frac{2.39724 \times 10^{-19}}{1.67 \times 10^{-27}}} \approx \sqrt{1.435385 \times 10^8} \approx 11800 \ \text{m/s} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Theory of Gases
The kinetic theory of gases offers an insightful perspective on understanding gas molecules' behavior. This theory suggests that gases are composed of many small particles, such as atoms or molecules, constantly moving in random directions.
Imagine these particles like tiny, fast-moving dots bouncing around in a jar. Their speed and movement are influenced by temperature. As the temperature rises, these particles move faster.
In our sun's example, the surface temperature is about 5800 K. Using the formula for root mean square (RMS) speed, \( v_{rms} = \sqrt{\frac{3kT}{m}} \), we calculate how quickly a hydrogen atom moves at this temperature.
  • \( k \) stands for the Boltzmann constant, a fundamental figure in physics that helps relate energy at the molecular scale to temperature.
  • \( T \) is the temperature, in this context, 5800 K, representing the sun's surface temperature.
  • \( m \) is the mass of a hydrogen atom, approximately \( 1.67 \times 10^{-27} \) kg.

Plugging these into the formula allows us to determine the hydrogen atom's RMS speed, resulting in an impressive \( 11800 \ m/s \).
This rapidly moving atom isn’t just part of the solar surface’s relentless dynamism—each molecule's movement is a tiny piece of the broader energetic puzzle of our sun.
Solar Physics
Solar physics delves into the study of the sun, our closest stellar neighbor and a crucial component of our solar system. The sun operates not only as a source of light and heat but also affects our entire solar system's environment.
Its surface temperature of about 5800 K means that even ordinary hydrogen, the simplest and most prevalent element in the universe, becomes extremely energetic.
  • The sun's energy output, fundamentally powered by nuclear fusion reactions in its core, radiates outward, influencing the solar atmosphere where these hydrogen atoms exist.
  • The extreme conditions on the sun are why many familiar earth-based concepts, like atmospheric escape, require adjusted understandings.

In these searing conditions, hydrogen atoms don't merely abide by earth-like behaviors. They become part of processes far beyond typical earthly physics, like high-energy solar winds and other complex solar phenomena.
Understanding solar physics enables scientists to unravel mysteries about solar activity like solar flares, sunspots, and the sun's magnetic field—all essential for predicting and understanding space weather.
Gravitational Escape Velocity
Gravitational escape velocity is a critical concept when considering whether particles, like hydrogen atoms on the sun, can escape the gravitational pull. This velocity refers to the minimum speed an object must reach to break free from the gravitational attraction of a celestial body without further propulsion.
For the sun, the formula to find this escape speed is \((2 G M / R)^{1 / 2},\) where
  • \( G \) is the gravitational constant, offering a measure of the strength of gravity.
  • \( M \) represents the mass of the sun, an immense figure given the sun's vastness.
  • \( R \) is the sun's radius.

Calculating this shows that hydrogen atoms on the sun's surface, traveling at their RMS speed of \( 11800 \ m/s \), do not reach the solar escape velocity, which is significantly higher.
Consequently, single hydrogen atoms don’t generally escape the sun simply due to their speed. However, solar phenomena like solar winds consist of particles accelerated by other forces, giving them an extra push off the sun’s surface into the solar system.
This distinction is crucial in explaining why, despite not having the escape speed on their own, hydrogen atoms can still move away from the sun influenced by external forces.

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Most popular questions from this chapter

Meteorology. The vapor pressure is the pressure of the vapor phase of a substance when it is in equilibrium with the solid or liquid phase of the substance. The relative humidity is the partial pressure of water vapor in the air divided by the vapor pressure of water at that same temperature, expressed as a percentage. The air is saturated when the humidity is 100\(\%\) (a) The vapor pressure of water at \(20.0^{\circ} \mathrm{C}\) is \(2.34 \times 10^{3}\) Pa. If the air temperature is \(20.0^{\circ} \mathrm{C}\) and the relative humidity is \(60 \%,\) what is the partial pressure of water vapor in the atmosphere (that is, the pressure due to water vapor alone)? (b) Under the conditions of part (a), what is the mass of water in 1.00 \(\mathrm{m}^{3}\) of air? (The molar mass of water is 18.0 \(\mathrm{g} / \mathrm{mol}\) . Assume that water vapor can be treated as an ideal gas.)

(a) Compute the increase in gravitational potential energy for a nitrogen molecule (molar mass 28.0 \(\mathrm{g} / \mathrm{mol} )\) for an increase in elevation of 400 \(\mathrm{m}\) near the earth's surface. (b) At what temperature is this equal to the average kinetic energy of a nitrogen molecule? (c) Is it possible that a nitrogen molecule near sea level where \(T=15.0^{\circ} \mathrm{C}\) could rise to an altitude of 400 \(\mathrm{m} ?\) Is it likely that it could do so without hitting any other molecules along the way? Explain.

CP BIO The Bends. If deep-sea divers rise to the surface too quickly, nitrogen bubbles in their blood can expand and prove fatal. This phenomenon is known as the bends. If a scuba diver rises quickly from a depth of 25 \(\mathrm{m}\) in Lake Michigan (which is fresh water), what will be the volume at the surface of an \(\mathrm{N}_{2}\) bubble that occupied 1.0 \(\mathrm{mm}^{3}\) in his blood at the lower depth? Does it seem that this difference is large enough to be a problem? (Assume that the pressure difference is due only to the changing water pressure, not to any temperature difference, an assumption that is reasonable, since we are warm-blooded creatures.)

A flask contains a mixture of neon (Ne), krypton (Kr), and radon (Rn) gases. Compare (a) the average kinetic energies of the three types of atoms and (b) the root-mean-square speeds. (Hint: The periodic table in Appendix D shows the molar mass (in g/mol) of each element under the chemical symbol for that element.)

Gaseous Diffusion of Uranium. (a) A process called gaseous diffusion is often used to separate isotopes of uranium that is, atoms of the elements that have different masses, such as 235 \(\mathrm{U}\) and 238 \(\mathrm{U} .\) The only gaseous compound of uranium at ordinary temperatures is uranium hexafluoride, UF \(_{6}\) . Speculate on how 235 \(\mathrm{UF}_{6}\) and \(^{238} \mathrm{UF}_{6}\) molecules might be separated by diffusion. (b) The molar masses for \(^{235} \mathrm{UF}_{6}\) and 238 \(\mathrm{UF}_{6}\) molecules are 0.349 \(\mathrm{kg} / \mathrm{mol}\) and \(0.352 \mathrm{kg} / \mathrm{mol},\) respectively. If uranium hexafluoride acts as an ideal gas, what is the ratio of the root-meansquare speed of \(^{235} \mathrm{UF}_{6}\) molecules to that of \(^{238} \mathrm{UF}_{6}\) molecules if the temperature is uniform?

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