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Chapter 40: Problem 51

A hypothetical atom has energy levels uniformly separated by \(1.2 \mathrm{eV}\). At a temperature of \(2000 \mathrm{~K},\) what is the ratio of the number of atoms in the 13 th excited state to the number in the 11th excited state?

Short Answer

Expert verified
The ratio of the number of atoms in the 13th excited state to the 11th excited state is very close to zero.

Step by step solution

01

Understanding Energy Levels

The problem states that energy levels are separated by \(1.2 \text{ eV}\). Let's identify energies of the excited states: the 11th excited state has energy \( E_{11} = 11 \times 1.2 \text{ eV} \) and the 13th excited state has energy \( E_{13} = 13 \times 1.2 \text{ eV} \).
02

Using Boltzmann Distribution

The number of atoms in a energy state is given by the Boltzmann distribution. The ratio of the number of atoms in two different energy states is given by: \[ \frac{n_{13}}{n_{11}} = \frac{e^{-E_{13} / kT}}{e^{-E_{11} / kT}} = e^{-(E_{13} - E_{11}) / kT} \] where \(k\) is Boltzmann's constant \(8.617333262145 \times 10^{-5} \text{ eV/K}\).
03

Calculate Energy Difference

Calculate the energy difference between the two states: \( E_{13} - E_{11} = (13 \times 1.2 - 11 \times 1.2) \text{ eV} = 2.4 \text{ eV} \).
04

Compute the Exponent

Now calculate the exponent of the exponential function: \[ \frac{2.4 \text{ eV}}{8.617 \times 10^{-5} \text{ eV/K} \times 2000 \text{ K}} = \frac{2.4}{0.17234} \approx 13.92 \].
05

Final Calculation of Ratio

Finally, compute the ratio \( \frac{n_{13}}{n_{11}} = e^{-13.92} \). This calculates to a very small number since \(e^{-13.92}\) is close to zero.

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Key Concepts

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

Energy Levels
In quantum mechanics, energy levels are specific energies that a quantum system, such as an atom or a molecule, can have. Think of energy levels as stairs where each step corresponds to certain energy states. In the problem, we have energy levels separated uniformly by a value of \(1.2 \text{ eV}\). This means that the energy difference between any two consecutive energy states is constant.

For excited states like the 11th and 13th, which are beyond the ground state, calculations involve multiples of this basic energy gap.
  • The 11th excited state energy is \( E_{11} = 11 \times 1.2 \text{ eV} \).
  • The 13th excited state energy is \( E_{13} = 13 \times 1.2 \text{ eV} \).
This uniform separation allows us to calculate differences between states easily, crucial for further calculations like those using the Boltzmann distribution.
Excited States
Atoms can exist in different energy states. When they have more energy than the least possible energy (ground state), they are considered to be in an excited state. There are various excited states, each higher in energy than the last, which are reached by absorbing energy.

The given problem explores two such excited states: the 11th and 13th. Excited states are important as they help determine how atoms behave under different conditions:
  • An atom in the 11th excited state has moved up 11 energy "steps" from the ground state.
  • An atom in the 13th excited state is two steps higher than the 11th, having absorbed energy to reach that level.
This concept is essential in physics and chemistry for understanding reactions and behaviors of materials under energy changes.
Boltzmann's Constant
Boltzmann's constant \(k\) is a fundamental physical constant that describes the relationship between temperature and energy at the microscopic level. It is a crucial element in the Boltzmann distribution, which expresses how particles distribute themselves among various energy states at a given temperature.

The value of Boltzmann's constant is \(8.617333262145 \times 10^{-5} \text{ eV/K}\). It is used in the exponential factor of the Boltzmann distribution, showing its pivotal role:

The distribution formula \( e^{-(E_{13} - E_{11}) / kT} \) uses this constant to quantify the ratio of how likely an atom is to be in one state versus another at a temperature \(T\). Hence, it determines the probability of atoms occupying higher energy levels and aids in understanding thermal equilibrium and behavior of gases.

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Most popular questions from this chapter

How many electron states are in these subshells: (a) \(n=4\), \(\ell=3 ;(\mathrm{b}) n=3, \ell=1 ;(\mathrm{c}) n=4, \ell=1 ;(\mathrm{d}) n=2, \ell=0 ?\)

X rays are produced in an \(x\) -ray tube by electrons accelerated through an electric potential difference of \(50.0 \mathrm{kV}\). Let \(K_{0}\) be the kinetic energy of an electron at the end of the acceleration. The electron collides with a target nucleus (assume the nucleus remains stationary) and then has kinetic energy \(K_{1}=0.500 K_{0}\). (a) What wavelength is associated with the photon that is emitted? The electron collides with another target nucleus (assume it, too, remains stationary) and then has kinetic energy \(K_{2}=0.500 K_{1}\) (b) What wavelength is associated with the photon that is emitted?

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