Chapter 14: Problem 17
(For all of these problems, where necessary assume the normal ratio of total- to-selective extinction.) For an excitation temperature of \(100 \mathrm{K},\) what is the ratio of populations for the two levels in the \(21 \mathrm{cm}\) transition? (Take the statistical weight of the lower level to be 1 and the upper level to be \(3 .\))
Short Answer
Step by step solution
- Understand the Problem
- Equation for Population Ratio
- Substitute Values
- Calculate Exponent
- Population Ratio
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Boltzmann Equation
\[ \frac{n_i}{n_0} = \frac{g_i}{g_0} \times e^{-\frac{E_{ij}}{kT}} \]
Here:
- \(n_i\): Population of the higher energy state
- \(n_0\): Population of the ground state
- \(g_i\): Statistical weight of the higher energy state
- \(g_0\): Statistical weight of the ground state
- \(E_{ij}\): Energy difference between the two states
- \(k\): Boltzmann constant (\(1.381 \times 10^{-23} J/K\))
- \(T\): Absolute temperature (in Kelvin)
This equation informs us that the higher the energy difference \(E_{ij}\) between two states, the fewer particles will populate the higher energy state at a given temperature \(T\).
21 cm Transition
The 21 cm line is important in astrophysics as it is used to study a variety of cosmic features and phenomena:
- Mapping the distribution and density of neutral hydrogen in our galaxy and others
- Understanding the structure and dynamics of galaxies
- Tracing the evolution of matter throughout cosmic history
Astrophysics Energy Levels
Key aspects to understand about energy levels include:
- Electrons in higher energy levels are less stable and are more likely to drop to lower energy levels, releasing energy.
- Energy levels are often described by quantum numbers which ensure the unique configuration of electrons.
- The difference in energy between these levels determines the wavelength and frequency of the resultant radiation.
Population Ratio Calculation
The steps are:
- Convert the 21 cm wavelength to energy: \(E_{ij} = \frac{hc}{\text{wavelength}} = \frac{6.626 \times 10^{-34} × 3 \times 10^8}{21 \times 10^{-2}} = 5.7 \times 10^{-24} J\)
- Calculate the exponent: \(\frac{E_{ij}}{kT} = \frac{5.7 \times 10^{-24}}{1.381 \times 10^{-23} \times 100} = 0.04125\)
- Substitute these values into the Boltzmann equation: \(\frac{n_i}{n_0} = \frac{g_i}{g_0} \times e^{-\frac{E_{ij}}{kT}} = 3 \times e^{-0.04125} ≈ 3 \times 0.9596 ≈ 2.8788\)
This value denotes the ratio of particles in the higher energy state to the ground state at the given temperature.