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Q 6.15

Expert-verifiedFound in: Page 231

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

Suppose you have 10 atoms of weberium: 4 with energy 0 eV, 3 with energy 1 eV, 2 with energy 4 eV, and 1 with energy 6 eV.

(a) Compute the average energy of all your atoms, by adding up all their energies and dividing by 10.

(b) Compute the probability that one of your atoms chosen at random would have energy E, for each of the four values of E that occur.

(c) Compute the average energy again, using the formula $\overline{E}=\sum _{s}E\left(s\right)\mathcal{P}\left(s\right)$

Therefore,

The average energy is $\overline{E}=1.7\mathrm{eV}$

The probability is $P\left(0\mathrm{eV}\right)=\frac{4}{10}P\left(1\mathrm{eV}\right)=\frac{3}{10}P\left(4\mathrm{eV}\right)=\frac{2}{10}P\left(6\mathrm{eV}\right)=\frac{1}{10}$

The average energy using formula is $\overline{E}=1.7\mathrm{eV}$

10 atoms of weberium: 4 with energy 0 eV, 3 with energy 1 eV, 2 with energy 4 eV, and 1 with energy 6 eV.

**(a)** Assume we have ten weberium atoms, four of which have an energy of E1 = 0 eV, three of which have an energy of E2 =1 eV, two of which have an energy of E3 4 eV, and one of which has an energy of E4 = 6 eV; the average energy is given by:

$\overline{E}=\frac{\sum N{E}_{N}}{\sum N}$

Where ${E}_{N}$ is the energy level that occupied by N atoms

Substitute the values,

$\overline{E}=\frac{4\left(0\mathrm{eV}\right)+3(1.0\mathrm{eV})+2\left(4\mathrm{eV}\right)+1\left(6\mathrm{eV}\right)}{10}=1.7\mathrm{eV}\phantom{\rule{0ex}{0ex}}\overline{E}=1.7\mathrm{eV}$

**(b)** Each energy's probability is equal to the number of atoms with that energy divided by the total number of atoms, so:

$P\left(0\mathrm{eV}\right)=\frac{4}{10}P\left(1\mathrm{eV}\right)=\frac{3}{10}\phantom{\rule{0ex}{0ex}}P\left(4\mathrm{eV}\right)=\frac{2}{10}P\left(6\mathrm{eV}\right)=\frac{1}{10}$

The average energy is:

$\overline{E}=\sum E\left(s\right)P\left(s\right)$

The average energy of the system is:

$\overline{E}={E}_{1}P\left(0\mathrm{eV}\right)+{E}_{2}P\left(1\mathrm{eV}\right)+{E}_{3}P\left(4\mathrm{eV}\right)+{E}_{4}P\left(6\mathrm{eV}\right)$

Substitute from part (b):

$\overline{E}=\left(0\mathrm{eV}\right)(0.4)+\left(1\mathrm{eV}\right)(0.3)+\left(4\mathrm{eV}\right)(0.2)+\left(6\mathrm{eV}\right)(0.1)\phantom{\rule{0ex}{0ex}}\overline{E}=1.7\mathrm{eV}$

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