Crystal Binding Questions and Answers

Answer: B

The Einstein frequency is \(\frac{1}{2\pi}\sqrt{\frac{\epsilon}{m e r_{0}^{2}}}\).

Each ion vibrates about the minimum of \(U(r)\) with spring constant \(U''\) there. With \(x = r/r_{0}\), \(U = -\epsilon x e^{-x}\):

\[ \begin{aligned} \frac{dU}{dx} &= -\epsilon(1 - x)e^{-x} = 0 \Rightarrow x = 1 \\ \frac{d^{2}U}{dx^{2}} &= \epsilon(2 - x)e^{-x} = \frac{\epsilon}{e} \text{ at } x = 1 \end{aligned} \]

So \(U''(r_{0}) = \epsilon/(e r_{0}^{2})\), and

\[ \begin{aligned} \nu_{E} &= \frac{1}{2\pi}\sqrt{\frac{U''}{m}} \\ &= \frac{1}{2\pi}\sqrt{\frac{\epsilon}{m e r_{0}^{2}}} \end{aligned} \]

The options with \(e\) in the numerator have the exponential factor inverted, and the factor 2 in the other options comes from evaluating \(2 - x\) at \(x = 0\) instead of at the minimum.

Answer: A

The spring constant is 156A − 42B.

In the harmonic limit the spring constant is the curvature of the potential at equilibrium, \( k = U''(r_{0}) \):

\[ \begin{aligned} U'(r) &= -\frac{12A}{r^{13}} + \frac{6B}{r^{7}} \\ U''(r) &= \frac{156A}{r^{14}} - \frac{42B}{r^{8}} \end{aligned} \]

Setting \( r_{0} = 1 \) gives \( 156A - 42B \). The forms in 12 and 6 are the first derivative, which is the force (and vanishes at equilibrium), not the stiffness, and 42A − 156B swaps the coefficients.