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Q51P

Expert-verifiedFound in: Page 1249

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 $9\times {10}^{-7}$.

The energy difference between two atomic levels of an atom, $\u2206\mathrm{E}=1.2\mathrm{eV}$

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

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

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, 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, $k=8.625\times {10}^{-5}\mathrm{eV}/\mathrm{K}$

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 13^{th} excited state and the 11^{th} 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 13^{th} excited state to the number of atoms in the 11^{th} excited state as follows:

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

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

**Here are the ${{\mathbf{K}}}_{{\mathbf{\alpha}}}$ wavelengths of a few elements:**

Element | λ (pm) | Element | λ (pm) |

Ti | 275 | Co | 179 |

V | 250 | Ni | 166 |

Cr | 229 | Cu | 154 |

Mn | 210 | Zn | 143 |

Fe | 193 | Ga | 134 |

**Make a Moseley plot (like that in Fig. 40-16) from these data and verify that its slope agrees with the value given for C in Module 40-6.**

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