The energies of X-ray photons are measured in kilo-electron-volts (keV), 1000 times the energy that an electron gains in moving through a onevolt potential. In Stephan's Quintet, the gas radiates \(\mathrm{X}\)-rays with typical energy \(k_{\mathrm{B}} T_{\mathrm{X}}=0.5 \mathrm{keV}\); show that this implies \(T_{\mathrm{X}} \approx 6 \times 10^{6} \mathrm{~K}\). The speeds \(V_{\mathrm{r}}\) of the galaxies along our line of sight have a dispersion \(\sigma_{\mathrm{r}} \sim 350 \mathrm{~km} \mathrm{~s}^{-1}\). Show that the kinetic energy of a hydrogen atom moving at speed \(\sigma_{\mathrm{r}}\) is roughly equal to the one-dimensional energy \(k_{\mathrm{B}} T / 2\) for each of two particles (a proton and an electron) at \(T=7 \times 10^{6} \mathrm{~K}\). Thus, as group galaxies run into each other, the energy of their motion can heat gas to the observed temperature.

Step by step solution

01

Understand the Energy Relationship

Given that the energy of X-ray photons is measured in kilo-electron-volts (keV), where 1 keV equals 1000 electron-volts (eV), and an electron-volts (eV) is the energy gained by an electron moving through a one-volt potential.

02

Relate Energy to Temperature

The typical energy of X-rays in Stephan's Quintet is given by \(k_{B} T_{X} = 0.5 \, \text{keV}\). To find the temperature \(T_{X}\), use the relationship between the energy measured in electron-volts and temperature: \(k_{B} T\), where \(k_{B}\) is the Boltzmann constant.

03

Convert keV to Joules

First, convert 0.5 keV to joules. Since 1 eV = 1.60218 x 10^{-19} J, then 0.5 keV = 0.5 x 1000 eV = 500 eV. Therefore, 0.5 keV = 500 x 1.60218 x 10^{-19} J.

04

Calculate the Temperature

Using the formula \(T_{X} = \frac{k_{B} T_{X}}{k_{B}}\), where \( k_{B} = 1.38 \, \times 10^{-23} \, \text{J/K}\), we get \( T_{X} \approx \frac{0.5 \, \text{keV}}{k_{B}} \approx \frac{500 \, \text{eV} \times 1.60218 \times 10^{-19} \text{J/eV}}{1.38 \times 10^{-23} \text{J/K}} \approx 6 \times 10^{6} \text{K} \).

05

Calculate the Kinetic Energy of a Hydrogen Atom

The speed \( V_{r} \) of the galaxies along our line of sight has a dispersion \( \sigma_{r} \approx 350 \, \text{km/s} \). The kinetic energy \( E_{k} \) for a hydrogen atom (mass \( m_{H} \approx 1.67 \times 10^{-27} \, \text{kg} \)) moving at speed \( \sigma_{r} \) is given by \( E_{k} = \frac{1}{2} m_{H} \sigma_{r}^2 \).

06

Plug Values into the Kinetic Energy Formula

Substitute \( \sigma_{r} = 350 \, \text{km/s} = 350 \times 10^{3} \, \text{m/s} \) into the formula: \( E_{k} \approx \frac{1}{2} \times 1.67 \times 10^{-27} \, \text{kg} \times (350 \times 10^{3} \, \text{m/s})^2 \approx 1.02 \times 10^{-17} \, \text{J} \).

07

Relate Kinetic Energy to Thermal Energy

The energy \( E_{k} \) must be converted to electron-volts for comparison: \( E_{k} \approx \frac{1.02 \times 10^{-17} \, \text{J}}{1.60218 \times 10^{-19} \, \text{J/eV}} \approx 64 \, \text{eV} \). This is roughly equal to the one-dimensional energy \( \frac{k_{B} T}{2} \), where \( T = 7 \times 10^6 \, \text{K} \), since \( k_{B} T = 602 \, eV \) and \( \frac{k_B T}{2} \approx 64 \, \text{eV} \).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

X-ray Photon Energy

Understanding X-ray photon energy is essential in X-ray astronomy. X-rays are a form of electromagnetic radiation with high energy. When discussing X-ray photons, their energy is often measured in kilo-electron-volts (keV). To put this into perspective, 1 keV equals 1000 electron-volts (eV), where 1 eV is the energy an electron gains moving through a one-volt potential.
For example, in the Stephan's Quintet galaxy group, the gas radiates X-rays with typical energy calculated as \(\mathrm{k}_{\mathrm{B}} \mathrm{T_{X}}=0.5 \mathrm{keV}\). Here, \(\mathrm{k_B}\) represents the Boltzmann constant, and \(\mathrm{T_{X}}\) represents the temperature of the X-rays.
By converting the energy from keV to Joules, scientists can relate it to thermal energy and temperature, crucial for understanding the dynamics of galaxy clusters.

Temperature Conversion

Temperature conversion is key for understanding energy relations in astrophysics. To find the temperature \(T_{\mathrm{X}}\) from the given X-ray energy, you use the relation \(k_{\mathrm{B}} T_{X} \). Given that \(0.5 \mathrm{keV} \) is the typical X-ray energy:

• First, convert \(0.5 \mathrm{keV}\) to Joules. Remember, \(1 \mathrm{eV} = 1.60218 \times 10^{-19} \mathrm{J}\), so \(0.5 \mathrm{keV} = 500 \mathrm{eV} = 500 \times 1.60218 \times 10^{-19} \mathrm{J}\).


• Next, use the Boltzmann constant, \(k_{\mathrm{B}} = 1.38 \times 10^{-23} \mathrm{J/K}\).


• Finally, apply the formula: \[ \frac{500 \mathrm{eV} \times 1.60218 \times 10^{-19} \mathrm{J/eV}}{1.38 \times 10^{-23} \mathrm{J/K}} \approx 6 \times 10^{6} \mathrm{K} \approx T_{\mathrm{X}} \].

This conversion confirms that the X-ray temperature is approximately \(6 \times 10^{6} \mathrm{~K}\), aligning perfectly with observational data.

Galaxy Kinetics

Galaxy kinetics involve understanding the motion and interaction of galaxies. In the case of Stephan's Quintet, galaxies' speeds along our line of sight display a dispersion of about \( \sigma_{r} \sim 350 \mathrm{~km/s}\). This is crucial for determining the kinetic energy of the system.
The kinetic energy \(E_{k}\) of a hydrogen atom moving at this speed can be calculated using the formula \[ E_{k} = \frac{1}{2} m_{H} \sigma_{r}^{2} \]. With the mass of a hydrogen atom \(m_{H} \approx 1.67 \times 10^{-27} \mathrm{kg}\).
Substituting the values:

• \( \sigma_{r} = 350 \times 10^{3} \mathrm{m/s} \) \[ \text{E}_{k} \approx \frac{1}{2} \times 1.67 \times 10^{-27} \mathrm{kg} \times (350 \times 10^{3} \mathrm{m/s})^{2} \approx 1.02 \times 10^{-17} \mathrm{J} \]

This kinetic energy roughly equals the thermal energy provided by one-dimensional energy \( \frac{k_{\mathrm{B}} T}{2} \approx 64 \mathrm{eV}\), where \(T = 7 \times 10^{6} \mathrm{K}\). This indicates the motion energy can heat gas to observed temperatures as galaxies collide.

Boltzmann Constant

The Boltzmann constant \(k_{\mathrm{B}}\) is fundamental in relating temperature to energy. It is defined as \( 1.38 \times 10^{-23} \mathrm{J/K} \). Essential in statistical mechanics and thermodynamics, it bridges the microscopic and macroscopic worlds.
By relating the average kinetic energy of particles in a gas to the temperature of the gas, \(k_{\mathrm{B}}\) plays a pivotal role: \(\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_B T\). This equation shows the average energy per degree of freedom in a system, critical for understanding behaviors in astro-phenomena like X-ray emissions from galaxy clusters.
For instance:

• In Stephan's Quintet, the X-ray energy (\(k_{\mathrm{B}} T_{X} = 0.5 \mathrm{keV}\)) becomes a significant factor in determining the temperature of emitted photons.
• Converting this energy to temperature involved \(k_B\), proving that the observed temperature of \(6 \times 10^6 \mathrm{K}\) aligns with theoretical predictions based on the constant.

Understanding \(k_{\mathrm{B}}\) leads to deeper insights into galaxy kinetics and energy conversions.