91Ó°ÊÓ

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:

$\stackrel{¯}{E}=\frac{∑N{E}_N}{∑N}$

Where $E_N$is the energy level that occupied by N atoms

Substitute the values,

$$\begin{gathered} \stackrel{¯}{E}=\frac{4\left(0\mathrm{eV}\right)+3(1.0\mathrm{eV})+2\left(4\mathrm{eV}\right)+1\left(6\mathrm{eV}\right)}{10}=1.7\mathrm{eV} \\ \stackrel{¯}{E}=1.7\mathrm{eV} \end{gathered}$$

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

$$\begin{gathered} P\left(0\mathrm{eV}\right)=\frac{4}{10}P\left(1\mathrm{eV}\right)=\frac{3}{10} \\ P\left(4\mathrm{eV}\right)=\frac{2}{10}P\left(6\mathrm{eV}\right)=\frac{1}{10} \end{gathered}$$

The average energy is:

$\stackrel{¯}{E}=∑E\left(s\right)P\left(s\right)$

The average energy of the system is:

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

Substitute from part (b):

$$\begin{gathered} \stackrel{¯}{E}=\left(0\mathrm{eV}\right)(0.4)+\left(1\mathrm{eV}\right)(0.3)+\left(4\mathrm{eV}\right)(0.2)+\left(6\mathrm{eV}\right)(0.1) \\ \stackrel{¯}{E}=1.7\mathrm{eV} \end{gathered}$$