Crystal Binding Questions and Answers

Q: For a covalently bonded solid consisting of ions of mass \(m\), the binding potential can be assumed to be given by\[ U(r) = -\epsilon\left(\tfrac{r}{r_{0}}\right)e^{-r/r_{0}} , \]where \(\epsilon\) and \(r_{0}\) are positive constants. What is the Einstein frequency of the solid in Hz?

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.

Q: A rare gas inter-atomic potential is given by \( U(r) = \frac{A}{r^{12}} - \frac{B}{r^{6}} \), where A and B are material parameters. What is the spring constant for displacement of atoms in the harmonic limit, if the given equilibrium separation \( r_{0} \) is 1 au :

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.

Q: If the interatomic potential energy function can be expressed as\[ U(R) = -\frac{A}{R^{6}} + \frac{B}{R^{12}}, \]where A and B are constants, the atomic spacing \(R_{0}\) for which the potential energy is a minimum is given by :

The equilibrium spacing is \(R_{0} = (2B/A)^{1/6}\).The minimum of U(R) is where its derivative vanishes:\[ \begin{aligned} \frac{dU}{dR} &= \frac{6A}{R^{7}} - \frac{12B}{R^{13}} = 0 \\ R^{6} &= \frac{2B}{A} \end{aligned} \]so \(R_{0} = (2B/A)^{1/6}\). This is the Lennard-Jones form, with an attractive \(R^{-6}\) van der Waals term and a repulsive \(R^{-12}\) term. The other options do not have the dimension of a length; \(A^{2}/4B\) is the depth of the well, \(-U_{min}\).

For a covalently bonded solid consisting of ions of mass \(m\), the binding potential can be assumed to be given by

\[ U(r) = -\epsilon\left(\tfrac{r}{r_{0}}\right)e^{-r/r_{0}} , \]

where \(\epsilon\) and \(r_{0}\) are positive constants. What is the Einstein frequency of the solid in Hz?

A.\(\frac{1}{2\pi}\sqrt{\frac{\epsilon e}{m r_{0}^{2}}}\)
B.\(\frac{1}{2\pi}\sqrt{\frac{\epsilon}{m e r_{0}^{2}}}\)
C.\(\frac{1}{2\pi}\sqrt{\frac{2\epsilon}{m e r_{0}^{2}}}\)
D.\(\frac{1}{2\pi}\sqrt{\frac{\epsilon e}{2m r_{0}^{2}}}\)

Exam Year(s)

GATE Physics - 2023,

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.

Discuss in Forum

A rare gas inter-atomic potential is given by \( U(r) = \frac{A}{r^{12}} - \frac{B}{r^{6}} \), where A and B are material parameters. What is the spring constant for displacement of atoms in the harmonic limit, if the given equilibrium separation \( r_{0} \) is 1 au :

A.156A − 42B
B.42A − 156B
C.12A − 6B
D.− 12A + 6B

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

Discuss in Forum

If the interatomic potential energy function can be expressed as

\[ U(R) = -\frac{A}{R^{6}} + \frac{B}{R^{12}}, \]

where A and B are constants, the atomic spacing \(R_{0}\) for which the potential energy is a minimum is given by :

A.A/4B
B.\((2B/A)^{1/6}\)
C.\(A^{2}/4B\)
D.\(4B^{2}/A\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The equilibrium spacing is \(R_{0} = (2B/A)^{1/6}\).

The minimum of U(R) is where its derivative vanishes:

\[ \begin{aligned} \frac{dU}{dR} &= \frac{6A}{R^{7}} - \frac{12B}{R^{13}} = 0 \\ R^{6} &= \frac{2B}{A} \end{aligned} \]

so \(R_{0} = (2B/A)^{1/6}\). This is the Lennard-Jones form, with an attractive \(R^{-6}\) van der Waals term and a repulsive \(R^{-12}\) term. The other options do not have the dimension of a length; \(A^{2}/4B\) is the depth of the well, \(-U_{min}\).

Discuss in Forum

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