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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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

The ratio of the number of atoms in the ${13}^{\mathrm{th}}$ excited state to the number of atoms in the ${11}^{\mathrm{th}}$ excited state is $9×{10}^{-7}$.

See the step by step solution

## Step 1: The given data:

The energy difference between two atomic levels of an atom, $∆\mathrm{E}=1.2\mathrm{eV}$

The temperature of the states, $\mathrm{T}=2000\mathrm{K}$

Number of atoms in the higher-energy state, ${\mathrm{N}}_{1}=6.1×{10}^{13}/{\mathrm{cm}}^{3}$

Number of atoms in the lower-energy state, ${\mathrm{N}}_{2}=2.5×{10}^{15}/{\mathrm{cm}}^{3}$

## Step 2: Understanding the concept of Boltzmann distribution equation

The Boltzmann distribution is a probability function used in statistical physics to define the state of a particle system in terms of temperature and energy.

Using the Boltzmann-distribution equation which is the expression for the probability for stimulated emission of radiation to the probability for spontaneous emission of radiation under thermal equilibrium, we can get the required ratio of the number of atoms present in the excited state to that present in the excited state.

Formula:

The Boltzmann energy distribution equation,

$\frac{{\mathrm{N}}_{1}}{{\mathrm{N}}_{2}}={\mathrm{e}}^{-\left({\mathrm{E}}_{2}-{\mathrm{E}}_{2}\right)/\mathrm{kT}}$ ….. (1)

Here, Boltzmann constant, $k=8.625×{10}^{-5}\mathrm{eV}/\mathrm{K}$

## Step 3: Calculation of the ratio of number of atoms in the given states:

The energy difference between any two levels is,

${\mathrm{E}}_{13}-{\mathrm{E}}_{11}=2\left(1.2\mathrm{eV}\right)\phantom{\rule{0ex}{0ex}}=2.4\mathrm{eV}$

So, the energy difference between the 13th excited state and the 11th excited state is given as:

Now, using this energy difference value in the equation (i) with the given data, we can get the ratio of the number of atoms in the 13th excited state to the number of atoms in the 11th excited state as follows:

$\frac{{\mathrm{N}}_{1}}{{\mathrm{N}}_{2}}={\mathrm{e}}^{-\frac{\left(2.4\mathrm{eV}\right)}{\left(8.625×{10}^{-5}\mathrm{eV}/\mathrm{K}\right)\left(2000\mathrm{K}\right)}}\phantom{\rule{0ex}{0ex}}={\mathrm{e}}^{-13.91}\phantom{\rule{0ex}{0ex}}=9×{10}^{-7}$

Hence, the value of the required ratio is $9×{10}^{-7}$.