Physics 4311: Thermal Physics – HW Solution 2

Problem 1: Boltzmann factors (10 points)

a)

Room temperature \(T= 300\) K, \(k_B T = 4.14 \times 10^{-21}\) J

b)

1 eV = \(1.602 \times 10^{-19}\) J, \(4.14 \times 10^{-21}\) J = 0.0258 eV

c)

Ionization energy \(E_{ion}=13.6\) eV, Boltzmann factor \(\exp {(-E_{ion}/k_B T)} = \exp {(-527)}\)
\(\Rightarrow \) hydrogen not appreciably ionized at room temperature

d)

Temperature at which hydrogen starts to ionize appreciably \(T_{ion} = E_{ion} /k_B = 1.59 \times 10^{5}\) K

e)

\(10^{-4}\) eV \(\ll k_B T\) \(\Rightarrow \) rotational energy levels will be excited

Impurity in solid (30 points)

a)

\[p_1 = \frac {e^{-E_1/k_BT}}{e^{-E_1/k_BT}+e^{-E_2/k_BT}} = \frac {e^{\epsilon /k_BT}}{e^{\epsilon /k_BT}+e^{-3\epsilon /k_BT}} = \frac {e^{2\epsilon /k_BT}}{e^{2\epsilon /k_BT}+e^{-2\epsilon /k_BT}}\]

b)

\[p_2 = \frac {e^{-E_2/k_BT}}{e^{-E_1/k_BT}+e^{-E_2/k_BT}} = \frac {e^{-3\epsilon /k_BT}}{e^{\epsilon /k_BT}+e^{-3\epsilon /k_BT}} = \frac {e^{-2\epsilon /k_BT}}{e^{2\epsilon /k_BT}+e^{-2\epsilon /k_BT}}\]

c)

\begin {eqnarray*} \langle E \rangle &=& p_1 E_1 + p_2 E_2 = -\epsilon p_1 + 3 \epsilon p_2 = \epsilon - 2\epsilon p_1 + 2\epsilon p_2 \\ &=& \epsilon - 2\epsilon \frac {e^{2\epsilon /k_BT}-e^{-2\epsilon /k_BT}}{e^{2\epsilon /k_BT}+e^{-2\epsilon /k_BT}} = \epsilon - 2\epsilon {\tanh (2\epsilon /k_B T)} \end {eqnarray*}

d)

\(T\to 0\): \(\qquad \epsilon /k_BT \to \infty \qquad \Rightarrow \langle E \rangle = \epsilon - 2\epsilon \tanh (\infty ) = -\epsilon \)
  \(T\to \infty \): \(\qquad \epsilon /k_BT \to 0 \qquad \Rightarrow \langle E \rangle = \epsilon - 2\epsilon \tanh (0) = \epsilon \)

e)

\[ C = \frac d {dT} \langle E \rangle = -2\epsilon \frac d {dT} \tanh (2\epsilon /k_B T) -2 \epsilon \frac 1 {\cosh ^2(2\epsilon /k_B T)} \frac {-2\epsilon }{k_B T^2} = k_B \left ( \frac {2\epsilon }{k_B T}\right )^2 \frac 1 {\cosh ^2(2\epsilon /k_B T)} \]

f)

\(T\to 0\): \(\qquad \epsilon /k_BT \to \infty ~, \qquad x^2/\cosh ^2(x) \to 0 \) for \(x\to \infty ~, \qquad \Rightarrow C=0\)
  \(T\to \infty \): \(\qquad \epsilon /k_BT \to 0 ~, \qquad x^2/\cosh ^2(x) \to 0 \) for \(x\to 0 ~, \qquad \Rightarrow C=0\)

g)

Find maximum of \(f(x)= x^2/\cosh ^2(x)\):

\begin {eqnarray*} \frac {df}{dx} &=& \frac {2x}{\cosh ^2(x)} - \frac {2x^2}{\cosh ^3(x)} \sinh (x) = 0 \\ 1 &=& x \tanh (x) \end {eqnarray*}

numerical solution: \(x_{max} = 1.1997 \qquad k_B T_{max} = 2\epsilon / 1.1997 = 1.667 \epsilon \)

h)