Question

Example 9.2 obtains a ratio of the number of particles expected in the \(n=2\) state to that in the ground state. Rather than the \(n=2\) state, consider arbitrary \(n\). (a) Show that the ratio is \(\frac{\text { number of energy } E_{n}}{\text { number of energy } E_{1}}=n^{2} e^{-13.6 \mathrm{cV}\left(1-n^{-2}\right) / k_{\mathrm{B}} T}\) Note that hydrogen atom energies are \(E_{n}=\) \(-13.6 \mathrm{eV} / \mathrm{r}^{2}\) (b) What is the limit of this ratio as \(n\) becomes very large? Can it exceed \(1 ?\) If so, under what condition(s)? (c) In Example 9.2, we found that even at the temperature of the Sun's surface \((\sim 6000 \mathrm{~K})\), the ratio for \(n=2\) is only \(10^{-8}\). For what value of \(n\) would the ratio be \(0.01 ?\) (d) Is it realistic that the number of atoms with high \(n\) could be greater than the number with low \(n\) ?

Short answer

a) The ratio of the number of particles in state \( n \) to the number of particles in the ground state is calculated with \( n^{2} e^{-13.6 (1-n^{-2}) / k_{B} T} \) where \( k_{B} \) is the Boltzmann constant and T is the temperature. b) As \( n \) becomes very large, the ratio will approach 0 and will not exceed 1. c) The value of \( n \) for which the ratio equals \( 0.01 \) can be calculated by using the given formula for \( T = 6000K \). d) The probability of having high \( n \) is less than having low \( n \) because of the exponential decrease.

Step-by-step solution

Evaluate the given formula

The ratio of the number of energy \( E_{n} \) to the number of energy \( E_{1} \) can be evaluated with \( n^{2} e^{-13.6 (1-n^{-2}) / k_{B} T} \), where \( k_{B} \) is the Boltzmann constant and T is the temperature.

Find the limit as \( n \) becomes very large

If \( n \) becomes very large, \( n^{-2} \) goes towards 0 and the ratio will approach 0 because the exponential factor goes to 0. This means the ratio couldn't exceed 1.

Identify the value of \( n \) when the ratio is 0.01

The value of \( n \) for which the ratio is \( 0.01 \) can be found by solving the given equation for \( n \), when \( T = 6000K \) and the ratio is \( 0.01 \).

Discuss if the number of atoms with high \( n \) could be greater than the number with low \( n \)

From the given ratio formula, we can see that as \( n \) increases, \( E_n \) increases, and the exponential term decreases, because the energy increases. This means the probability of having high \( n \) is less than having low \( n \).

Key concepts

Energy Level Ratios

When studying quantum mechanics and specifically the hydrogen atom, one often encounters the scenario of comparing the population of electrons in different energy states. The ratio of the number of particles in an excited state, represented by energy level 'n', to those in the ground state (the first energy level) is an important concept for understanding atomic behavior.

To derive this ratio, we use the formula provided in the exercise: \[\begin{equation}\frac{\text{number of energy } E_{n}}{\text{number of energy } E_{1}} = n^{2} e^{-13.6 \text{eV}(1-n^{-2}) / k_{B} T}\end{equation}\]Where 'n' is the principal quantum number, 'E' represents energy, 'k_{B}' is the Boltzmann constant, and 'T' is the temperature in Kelvin. As 'n' increases, the formula considers the increased probability due to the degeneracy factor ('n^2'), against the declining probability due to the exponential term influenced by the increased energy level and temperature.

Understanding this delicate balance is crucial for predicting electron distribution in quantum systems.

Boltzmann Distribution

The Boltzmann distribution plays a pivotal role in statistical mechanics, describing the distribution of particles over various energy states in thermal equilibrium. It's expressed mathematically as: \[\begin{equation} P(E) = e^{-E / (k_{B} T)}\end{equation}\]Type of writing: Friendly and super simpleWhere 'P(E)' is the probability of finding a particle in a state with energy 'E', 'k_{B}' is Boltzmann's constant, and 'T' is the absolute temperature. The negative exponent indicates that higher energy states are less likely to be populated than lower ones.

In the context of the hydrogen atom, the Boltzmann distribution can be used to explain why, at a given temperature, electrons are more likely to be found in lower energy states than higher ones. The exercise applies the Boltzmann principle to derive the ratio of electron populations between different energy levels of hydrogen, showing the interplay between energy considerations and statistical probabilities.

Hydrogen Atom Quantum States

Quantum states of the hydrogen atom are characterized by several quantum numbers, with the principal quantum number 'n' defining the energy level of the electron. The energy of an electron in a hydrogen atom is inversely proportional to the square of 'n' and can be represented as: \[\begin{equation}E_{n} = -\frac{13.6 \text{eV}}{n^{2}}\end{equation}\]Type of writing: Friendly and super simpleThe negative sign indicates that these energy levels are bound states, meaning the electron is attached to the nucleus. As 'n' increases, the electron moves to higher energy levels, which are further away from the nucleus, and the energy becomes less negative (i.e., closer to zero). This relationship is essential for understanding electron transitions and the emission or absorption of photons which correspond to specific wavelengths of light.

Limit of Quantum State Ratio

Considering the behavior of the ratio between quantum states as 'n' approaches infinity leads to a fundamental insight into quantum mechanics and the limitations of electron excitation. As outlined in the exercise solution, the limit can be expressed as: \[\begin{equation}\lim_{n \to \infty} \frac{\text{number of energy } E_{n}}{\text{number of energy } E_{1}}\end{equation}\]Type of writing: Friendly and super simpleFor large values of 'n', the term '-13.6 \text{eV}(1-n^{-2})' trends towards 0, causing the exponential term to approach 1, and since the prefactor 'n^{2}' increases without bound, one might expect the ratio to increase indefinitely. However, due to the properties of the exponential function, the ratio does not exceed 1, showing a physical limit to the population of higher energy states even at very high 'n' values. This illustrates the practical implications of quantum mechanics—while theoretically possible, in reality, higher energy states become increasingly less populated.