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An Introduction to Thermal Physics

Daniel V. Schroeder

Physics

1st Edition 路 356 pages

ISBN: 9780201380279

Chapter 8: Q. 8.9 (page 338)

Show that the Lennard-Jones potential reaches its minimum value at $r = r_{0}$ , and that its value at this minimum is $- u_{0}$ . At what value of $r$ does the potential equal zero?

Short Answer

Expert verified

At, $r = \frac{r_{0}}{2^{\frac{1}{6}}}$ the value of Lennard-Jones potential become 0.

Step by step solution

01

Step 1. Given information

Lennard- Jones potential=

$u (r) = u_{0} [{(\frac{r_{0}}{r})}^{12} - 2 {(\frac{r_{0}}{r})}^{6}]$

02

Step 2. Differentiating both sides with respect to r

We get,

$\frac{d u}{d r} = u_{0} [r_{0}^{12} 脳 - 12 \frac{1}{r^{13}} - 2 r_{0}^{6} 脳 - 6 \frac{1}{r^{7}}]$

= u_{0} [- 12 \frac{r_{0}^{12}}{r^{13}} + 12 \frac{r_{0}^{6}}{r^{7}}]

For extreme points,

$\frac{d u}{d r} = 0$

$鈭� u_{0} [- 12 \frac{r_{0}^{12}}{r^{13}} + 12 \frac{r_{0}^{6}}{r^{7}}] = 0$

$\frac{12 r_{0}^{6} u_{0}}{r^{13}} [r^{6} - r_{0}^{6}] = 0$

$r^{6} - r_{0}^{6} = 0$

$r = r_{0}$

Thus the Lennard-Jones potential has extremum point at $r = r_{0}$

03

Step 3. To check that the potential at r=r0 is maximum or minimum take double derivative

We get,

\frac{d^{2} u}{d r^{2}} = u_{0} [- 12 脳 \frac{- 13 r_{0}^{12}}{r^{14}} + 12 脳 \frac{- 7 r_{0}^{6}}{r^{8}}]

$= 12 u_{0} [\frac{13 r_{0}^{12}}{r^{14}} - \frac{7 r_{0}^{6}}{r^{8}}]$

${\frac{d^{2} u}{d r^{2}}|}_{r = s_{0}} = 12 u_{0} [\frac{13 r_{0}^{12}}{r_{0}^{14}} - \frac{7 r_{0}^{6}}{r_{0}^{8}}]$

$= 12 u_{0} [\frac{13}{r_{0}^{2}} - \frac{7}{r_{0}^{2}}]$

$= \frac{12 u_{0}}{r_{0}^{2}} [6]$

$= \frac{72 u_{0}}{r_{0}^{2}} > 0$

${鈭� \frac{d^{2} u}{d r^{2}}|}_{r = r_{0}} > 0$

So the Lennard- Jones potential has minimum value at $r^{'} = r_{0}$

${u (r)|}_{r = r_{0}} = u_{0} {[{(\frac{r_{0}}{r})}^{12} - 2 {(\frac{r_{0}}{r})}^{6}]}_{r = r_{0}}$

$= u_{0} [{(\frac{r_{0}}{r})}^{12} - 2 {(\frac{r_{0}}{r})}^{6}]$

$= u_{0} [1 - 2]$

${u (r)|}_{r = r_{0}} = - u_{0}$

04

Step 4. If the potential becomes 0.

$u (r) = 0$

$u_{0} [{(\frac{r_{0}^{*}}{r})}^{12} - 2 {(\frac{r_{0}}{r})}^{6}] = 0$

$\frac{u_{0}}{r^{12}} [r_{0}^{12} - 2 r_{0}^{6} r^{6}] = 0$

$r_{0}^{12} - 2 r_{0}^{6} r^{6} = 0$

$r^{6} = \frac{r_{0}^{12}}{2 r_{0}^{6}} = \frac{r_{0}^{6}}{2}$

$r^{6} = \frac{r_{0}^{12}}{2 r_{0}^{6}}$

$r = \frac{1}{2^{\frac{1}{6}}} r_{0}$

Thus the Lennard-Jones potential becomes zero at $r = \frac{1}{2^{\frac{1}{6}}} r_{0}$

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Most popular questions from this chapter

In this problem you will use the mean field approximation to analyse the behaviour of the Ising model near the critical point.

(a) Prove that, when $x 鈮� 1, \tanh x 鈮� x - \frac{1}{3} x^{3}$

(b) Use the result of part (a) to find an expression for the magnetisation of the Ising model, in the mean field approximation, when T is very close to the critical temperature. You should find $M 鈭� {(T_{c} - T)}^{\overset{炉}{尾}}, where 尾$ (not to be confused with 1/kT) is a critical exponent, analogous to the f defined for a fluid in Problem 5.55. Onsager's exact solution shows that $尾 = 1 / 8$ in two dimensions, while experiments and more sophisticated approximations show that $尾鈮� 1 / 3$ in three dimensions. The mean field approximation, however, predicts a larger value.

(c) The magnetic susceptibility $蠂$ is defined as $蠂鈮� (鈭� M / 鈭� B)_{T}$ . The behaviour of this quantity near the critical point is conventionally written as $蠂鈭� {(T - T_{c})}^{- 纬}$ , where y is another critical exponent. Find the value of in the mean field approximation, and show that it does not depend on whether T is slightly above or slightly below Te. (The exact value of y in two dimensions turns out to be 7/4, while in three dimensions $纬鈮� 1.24$ .)

Starting from the partition function, calculate the average energy of the one-dimensional Ising model, to verify equation 8.44. Sketch the average energy as a function of temperature.

Consider an Ising model in the presence of an external magnetic field B, which gives each dipole an additional energy of $- 渭_{B}$ B if it points up and $+ 渭_{B}$ B if it points down (where $渭_{B}$ is the dipole's magnetic moment). Analyse this system using the mean field approximation to find the analogue of equation 8.50. Study the solutions of the equation graphically, and discuss the magnetisation of this system as a function of both the external field strength and the temperature. Sketch the region in the T-B plane for which the equation has three solutions.

Consider a gas of "hard spheres," which do not interact at all unless their separation distance is less than $r_{0}$ , in which case their interaction energy is infinite. Sketch the Mayer $f$ -function for this gas, and compute the second virial coefficient. Discuss the result briefly.

For a two-dimensional Ising model on a square lattice, each dipole (except on the edges) has four "neighbors"-above, below, left, and right. (Diagonal neighbors are normally not included.) What is the total energy (in terms of $蔚$ ) for the particular state of the $4 脳 4$ square lattice shown in Figure 8.4?

Figure 8.4. One particular state of an Ising model on a $4 脳 4$ square lattice (Problem 8.15).

See all solutions

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