/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 \(\bullet$$\bullet\) (a) A deute... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\bullet$$\bullet\) (a) A deuteron, \(_{1}^{2} \mathrm{H},\) is the nucleus of a hydrogen isotope and consists of one proton and one neutron. The plasma of deuterons in a nuclear fusion reactor must be heated to about 300 million K. What is the rms speed of the deuterons? Is this a significant fraction of the speed of light \(\left(c=3.0 \times 10^{8} \mathrm{m} / \mathrm{s}\right) ?(\mathrm{b})\) What would the temperature of the plasma be if the deuterons had an rms speed equal to 0.10\(c ?\)

Short Answer

Expert verified
The RMS speed of deuterons at 300 million K is about 0.3% of the speed of light. Temperature needed for 0.10c is about 338 million K.

Step by step solution

01

Understanding Root Mean Square (RMS) Speed Formula

To find the root mean square (RMS) speed \( v_{\text{rms}} \) of deuterons, we use the formula: \( v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \), 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 deuteron. The mass of a deuteron \( m \) is approximately \( 3.34 \times 10^{-27} \text{ kg} \) (sum of the masses of a proton and a neutron).
02

Calculating RMS Speed at 300 Million K

Substitute the given temperature \( T = 300 \times 10^6 \text{ K} \) into the RMS speed formula: \[v_{\text{rms}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300 \times 10^6}{3.34 \times 10^{-27}}}\] Simplify and solve to find \( v_{\text{rms}} \).
03

Solving for RMS Speed

Carrying out the arithmetic, we find: \[v_{\text{rms}} \approx 9.1 \times 10^5 \text{ m/s}\] This is the RMS speed of deuterons at 300 million K.
04

Comparing RMS Speed to Speed of Light

The speed of light \( c = 3.0 \times 10^8 \text{ m/s} \). Compare \( v_{\text{rms}} \approx 9.1 \times 10^5 \text{ m/s} \) to \( c \). The fraction of the speed of light is: \[\frac{9.1 \times 10^5}{3.0 \times 10^8} \approx 0.003\] This means \( v_{\text{rms}} \) is about 0.3% of \( c \), which is small.
05

Determining Temperature for 0.10c Speed

To find the temperature where the RMS speed is \( 0.10c \), set \( v_{\text{rms}} = 0.10 \times 3.0 \times 10^8 \text{ m/s} = 3.0 \times 10^7 \text{ m/s} \). Substitute in the RMS speed formula: \[3.0 \times 10^7 = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times T}{3.34 \times 10^{-27}}}\] Solve for \( T \).
06

Calculating the Required Temperature

Square both sides of the equation to eliminate the square root, then solve for \( T \): \[T = \frac{(3.0 \times 10^7)^2 \times 3.34 \times 10^{-27}}{3 \times 1.38 \times 10^{-23}}\] This simplifies to: \[T \approx 3.38 \times 10^8 \text{ K}\] Thus, the temperature needed for deuterons to have an RMS speed of \( 0.10c \) is approximately 338 million K.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Root Mean Square Speed
Root mean square (RMS) speed is a method commonly used to quantify the speed of particles in a gas. It gives an average speed that accounts for both the direction and the magnitude of the particle's speed. In physics, the RMS speed is defined using the formula:
  • \( v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \)
In this formula, \( k \) represents the Boltzmann constant, \( T \) is the absolute temperature in Kelvin, and \( m \) is the mass of the particle. This equation helps us understand how speed varies with increasing temperature and different particle masses. The RMS speed is crucial in understanding how particles in environments like a nuclear fusion reactor move and interact.
Boltzmann Constant
The Boltzmann constant \( k = 1.38 \times 10^{-23} \text{ J/K} \) is a fundamental physical constant that relates the average kinetic energy of particles in a gas with the temperature of the gas. This constant is a bridge between macroscopic and microscopic physics, allowing us to link temperatures, a macroscopic concept, with the microscopic world of particles.The Boltzmann constant is pivotal in equations such as the root mean square speed formula. It effectively translates temperature into kinetic energy, illustrating how energy is distributed among particles in a system.
Speed of Light
The speed of light, denoted by \( c = 3.0 \times 10^8 \text{ m/s} \), is a fundamental constant of nature. It is the speed at which all electromagnetic waves travel through a vacuum. In the context of nuclear fusion and high-energy physics, comparing particle speeds to the speed of light gives insight into their relativistic effects.For example, in the case of deuterons with an RMS speed, it's useful to express this speed as a fraction of the speed of light. If deuterons reach a significant fraction of the speed of light, relativistic physics starts to play a role. This comparison helps us assess how close our system is to realms where special relativity becomes important.
Temperature Calculation
Calculating temperature in high-energy systems like those found in nuclear fusion reactors is crucial, as temperature directly affects particle speed and kinetic energy. When using the RMS speed equation, adjusting each variable allows us to solve for the desired temperature. Suppose we need to find the temperature such that the RMS speed reaches a particular value (e.g., a fraction of the speed of light). We rearrange the RMS speed formula, setting our desired RMS speed and solving for \( T \):
  • \( T = \frac{(v_{\text{rms}})^2 \times m}{3k} \)
This rearranged formula shows how particle speed dictates temperature requirements. By targeting specific temperatures, we can optimize conditions for various processes, such as achieving efficient nuclear fusion. Thus, understanding and calculating temperature is key to operating high-energy physics experiments successfully.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\bullet$$\bullet\) The effect of altitude on the lungs. (a) Calculate the change in air pressure you will experience if you climb a 1000 \(\mathrm{m}\) moun- tain, assuming that the temperature and air density do not change over this distance and that they were \(22^{\circ} \mathrm{Cand} 1.2 \mathrm{kg} / \mathrm{m}^{3},\) respectively, at the bottom of the mountain. (b) If you took a 0.50 \(\mathrm{L}\) breath at the foot of the mountain and managed to hold it until you reached the top, what would be the volume of this breath when you exhaled it there?

\(\bullet$$\bullet\) A bicyclist uses a tire pump whose cylinder is initially full of air at an absolute pressure of \(1.01 \times 10^{5}\) Pa. The length of stroke of the pump (the length of the cylinder) is 36.0 \(\mathrm{cm}\) . At what part of the stroke (i.e., what length of the air column) does air begin to enter a tire in which the gauge pressure is \(2.76 \times 10^{5} \mathrm{Pa} ?\) Assume that the temperature remains constant during the compression.

\(\bullet$$\bullet\) A flask with a volume of 1.50 \(\mathrm{L}\) , provided with a stopcock, contains ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) at 300 \(\mathrm{K}\) and atmospheric pressure \(\left(1.013 \times 10^{5} \mathrm{Pa}\right) .\) The molar mass of ethane is 30.1 \(\mathrm{g} / \mathrm{mol}\) . The system is warmed to a temperature of \(380 \mathrm{K},\) with the stopcock open to the atmosphere. The stopcock is then closed, and the flask is cooled to its original temperature. (a) What is the final pressure of the ethane in the flask? (b) How many grams of ethane remain in the flask?

\(\bullet\) An ideal gas expands while the pressure is kept constant. During this process, does heat flow into the gas or out of the gas? Justify your answer.

\(\bullet$$\bullet\) STP. The conditions of standard temperature and pressure (STP) are a temperature of \(0.00^{\circ} \mathrm{C},\) and a pressure of 1.00 atm. (a) How many liters does 1.00 mol of any ideal gas occupy at STP? (b) For a scientist on Venus, an absolute pressure of 1 Venusian-atmosphere is 92 Earth-atmospheres. Of course she would use the Venusian atmosphere to define STP. Assuming she kept the same temperature, how many liters would 1 mole of ideal gas occupy on Venus?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.