Question

A large number of neon atoms are in thermal equilibrium. What is the ratio of the number of atoms in a 5\(s\) state to the number in a 3\(p\) state at (a) 300 K; (b) 600 K; (c) 1200 K? The energies of these states, relative to the ground state, are E\(_{5s}\) = 20.66 eV and E\(_{3p}\) = 18.70 eV. (d) At any of these temperatures, the rate at which a neon gas will spontaneously emit 632.8-nm radiation is quite low. Explain why.

Short answer

At 300 K, the ratio is nearly zero. At 600 K, it increases slightly. At 1200 K, it's higher but still low. Spontaneous emission is low because of insufficient thermal energy and population.

Step-by-step solution

Understand the concept

The exercise involves calculating the ratio of neon atoms in different energy states. The relevant formula for the ratio of atoms in two energy states in thermal equilibrium is derived from the Boltzmann distribution.

Apply the Boltzmann distribution

Using the Boltzmann distribution, the ratio of the number of atoms in states \(s\) and \(p\) at temperature \(T\) is given by:\[\frac{N_{5s}}{N_{3p}} = \frac{g_{5s}}{g_{3p}} \exp\left(\frac{-(E_{5s} - E_{3p})}{kT}\right)\]where \(g\) are degeneracy factors, \(E\) are energies, and \(k\) is Boltzmann's constant \(8.6173 \times 10^{-5} \text{ eV/K}\). Assume the degeneracy factors \(g_{5s} = g_{3p} = 1\) if not specified otherwise.

Calculate for 300 K

First, calculate the ratio for \(T = 300 \text{ K}\):\[\frac{N_{5s}}{N_{3p}} = \exp\left(\frac{-(20.66 - 18.70)}{8.6173 \times 10^{-5} \times 300}\right)\]Compute the exponent:\[\frac{N_{5s}}{N_{3p}} = \exp(-76.94)\]This results in a very small value, indicating very few atoms are in the \(5s\) state compared to the \(3p\) state.

Calculate for 600 K

Repeat the calculation for \(T = 600 \text{ K}\):\[\frac{N_{5s}}{N_{3p}} = \exp\left(\frac{-(20.66 - 18.70)}{8.6173 \times 10^{-5} \times 600}\right)\]Calculate the exponent:\[\frac{N_{5s}}{N_{3p}} = \exp(-38.47)\]This is still a small value, but higher compared to 300 K.

Calculate for 1200 K

Do the calculation for \(T = 1200 \text{ K}\):\[\frac{N_{5s}}{N_{3p}} = \exp\left(\frac{-(20.66 - 18.70)}{8.6173 \times 10^{-5} \times 1200}\right)\]Compute the exponent:\[\frac{N_{5s}}{N_{3p}} = \exp(-19.23)\]This value is significantly higher than at 300 K and 600 K.

Discuss spontaneous emission at 632.8 nm

Spontaneous emission is low because it typically occurs on longer timescales and requires a stimulated emission process or significant thermal energy, both of which are less effective for low population states like \(5s\) at these temperatures.

Key concepts

Neon atoms

Neon atoms are fascinating elements which reside in the noble gas group on the periodic table. They are colorless, odorless gases under ordinary conditions but exhibit a distinct reddish-orange glow when exposed to high voltage in a vacuum tube, making them popular in neon signs. These inert atoms have complete outer electron shells, making them stable and largely non-reactive with other elements.

In physics, neon atoms are often studied in terms of their atomic structure and energy states. Each neon atom consists of a nucleus surrounded by electrons stacked in various energy levels or orbitals. These energy levels are central to how neon atoms interact with radiation and heat.

Because neon atoms have clearly defined energy levels, they are excellent subjects for studying how atoms transition between these energy states under different conditions, such as varying temperatures.

Energy states

Energy states refer to the different levels of energy that the electrons around an atom can have. In neon atoms, these states can be thought of as "steps" in an energy ladder. Electrons in neon can move between these steps by absorbing or releasing energy.

Each energy state can be uniquely identified by its energy value, for instance, given as \(E_{5s}\) and \(E_{3p}\) in the problem. Here, the 5s and 3p labels denote specific electron configurations in these higher energy levels. Each of these states possess a specific energy difference compared to a reference point, often the ground state.

These energy differences are crucial for predicting how many atoms occupy each state at a given temperature. The population of atoms in these energy states can be determined using statistical methods, such as the Boltzmann distribution, which is a key concept in statistical mechanics.

Thermal equilibrium

Thermal equilibrium is a state in which all parts of a system have reached a uniform temperature and no net energy flows between parts. In our neon atom scenario, thermal equilibrium means that the distribution of atoms across different energy states remains constant over time at each given temperature.

When neon atoms are in thermal equilibrium, their energy state populations are described by the Boltzmann distribution. This distribution shows that more atoms will occupy the lower energy states than higher ones, especially at lower temperatures. The ratio of atoms in different energy states at thermal equilibrium becomes a focus of study because it reveals important insights into atomic behavior and is dictated by variables such as temperature and energy differences between states.

• The lower the temperature, the fewer atoms will be in higher energy states.
• As temperature increases, more atoms can jump to higher energy levels.

Spontaneous emission

Spontaneous emission is a process where an excited electron in a higher energy state can "fall" to a lower energy state, releasing a photon in the process. This is a natural process that doesn't require external stimulus, unlike stimulated emission, which is what lasers are based on.

For neon, when an electron from a higher energy state such as \(5s\) spontaneously drops to a lower state like \(3p\), light is emitted. The study of this behavior helps in understanding the spectra of gases and is used in applications like gas discharge lamps.

However, as noted in the exercise, spontaneous emission rate is quite low at 632.8 nm. This low rate occurs because:

• There are fewer atoms in the higher \(5s\) state, especially at lower temperatures, reducing potential emission events.
• Transition rates for spontaneous processes are inherently slower compared to stimulated processes.
• Low thermal energies in typical conditions further decrease the likelihood of spontaneous emissions.

This aspect of spontaneous emission contributes to its selective application in specific conditions or technologies where emissions can be effectively harnessed.