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Q50P

Expert-verifiedFound in: Page 1249

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A hypothetical atom has only two atomic energy levels, separated by 32 eV . Suppose that at a certain altitude in the atmosphere of a star there are ${\mathbf{6}}{\mathbf{.}}{\mathbf{1}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{13}}}{\mathbf{}}{\mathbf{/}}{{\mathbf{cm}}}^{{\mathbf{3}}}$ of these atoms in the higher-energy state and ${\mathbf{2}}{\mathbf{.}}{\mathbf{5}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{15}}}{\mathbf{}}{\mathbf{/}}{{\mathbf{cm}}}^{{\mathbf{3}}}$ in the lower-energy state. What is the temperature of the star’s atmosphere at that altitude?**

The temperature of the star’s atmosphere at that altitude is $1\times {10}^{4}\mathrm{K}$.

The energy difference between two atomic levels of an atom, $\u2206\mathrm{E}=3.2\mathrm{eV}$

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

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

**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 defines the temperature of the star at that altitude using the given data in the probability equation.**

Formula:

The Boltzmann energy distribution equation,

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

Here, a constant, $k=1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$

Using the given data in equation (1), the temperature of the star’s atmosphere at that altitude is as follow.

$\mathrm{T}=\frac{{\mathrm{E}}_{2}-{\mathrm{E}}_{1}}{\mathrm{kIn}\left({\mathrm{N}}_{2}/{\mathrm{N}}_{1}\right)}$

Substitute known values in the above equation.

$\mathrm{T}=\frac{3.2\mathrm{eV}}{\left(1.38\times {10}^{-23}\right)\mathrm{In}\left(2.5\times {10}^{15}/6.1\times {10}^{13}\right)}\phantom{\rule{0ex}{0ex}}=1\times {10}^{4}\mathrm{K}$

Hence, the value of the temperature is $1\times {10}^{4}\mathrm{K}$ .

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