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2. Consider a classical particle moving in a one-dimensional potential well $u\left(x\right)$, as shown The particle is in thermal equilibrium with a reservoir at temperature $T$, so the probabilities of its various states are determined by Boltzmann statistics.

{a) Show that the average position of the particle is given by

where each integral is over the entire $x$axis.

A one-dimensional potential well. The higher the temperature, the farther the particle will stray from the equilibrium point.

(b) If the temperature is reasonably low (but still high enough for classical mechanics to apply), the particle will spend most of its time near the bottom of the potential well. In that case we can expand $u\left(x\right)$in a Taylor series about the equilibrium point $x_0$: $u\left(x\right)=u\left(x_0\right)+{\left.\left(x-x_0\right)\frac{du}{dx}\right|}_{x_0}+{\left.\frac{1}{2}{\left(x-x_0\right)}^2\frac{d^{2}u}{d{x}^2}\right|}_{x_0}$

$+{\left.\frac{1}{3!}{\left(x-x_0\right)}^3\frac{d^{3}u}{d{x}^3}\right|}_{x_0}+⋯$

Show that the linear term must be zero, and that truncating the series after the quadratic term results in the trivial prediction $x=x_0$.

(c) If we keep the cubic term in the Taylor series as well, the integrals in the formula for $x$become difficult. To simplify them, assume that the cubic term is small, so its exponential can be expanded in a Taylor series (leaving the quadratic term in the exponent). Keeping only the smallest temperature-dependent term, show that in this limit x differs from zo by a term proportional to $kT$. Express the coefficient of this term in terms of the coefficients of the Taylor series for$u\left(x\right)$

(d) The interaction of noble gas atoms can be modeled using the Lennard Jones potential,

$u\left(x\right)=u_0\left[{\left(\frac{x_0}{x}\right)}^{12}-2{\left(\frac{x_0}{x}\right)}^6\right]$

Sketch this function, and show that the minimum of the potential well is at $x=x_0$, with depth $u_0$. For argon, $x_0=3.9A$and $u_0=0.010eV$. Expand the Lennard-Jones potential in a Taylor series about the equilibrium point, and use the result of part ( c) to predict the linear thermal expansion coefficient of a noble gas crystal in terms of $u_0$. Evaluate the result numerically for argon, and compare to the measured value

$\mathrm{Î\pm }=0.0007{\mathrm{K}}^{-1}(\text{at}80\mathrm{K})$

Step by step solution

01

part(a) Step 1:Given information

Let the particle be associated with position

02

part(a) Step 2: Simplify

$\stackrel{¯}{x}=\underset{x}{∑}x\mathcal{P}\left(x\right)=\underset{x}{∑}x\frac{e^{-\mathrm{β}u\left(x\right)}}{Z}$

03

part(b) Step 1:Given information

The linear tem in the expansion is zero

04

part(b) Step 2: Simplify

$∫x{e}^{\left.-\mathrm{β}âˆ\pounds u\left(x_0\right)+a{\left(x-x_0\right)}^2\right)}dx=e^{-\mathrm{β}u\left(x_0\right)}∫x{e}^{-\mathrm{β}a{\left(x-x_0\right)}^2}dx=e^{-\mathrm{β}u\left(x_0\right)}∫\left(y+x_0\right)e^{-\mathrm{β}a{y}^2}dy$

$∫e^{-\mathrm{β}\left[u\left(x_0\right)+a{\left(x-x_0\right)}^2\right]}dx=e^{-\mathrm{β}u\left(x_0\right)}∫e^{-\mathrm{β}a{\left(x-x_0\right)}^2}dx=e^{-\mathrm{β}u\left(x_0\right)}∫e^{-\mathrm{β}a{y}^2}dy,$

05

part(c) Step 1:Given information

we use Boltzman factor

06

part(c) Step 2: Simplify

$\stackrel{¯}{x}=\frac{e^{-\mathrm{β}u\left(x_0\right)}∫\left(y+x_0\right)e^{-\mathrm{β}a{y}^2}\left[1-\mathrm{β}b{y}^3\right]dy}{e^{-\mathrm{β}u\left(x_0\right)}∫e^{-\mathrm{β}a{y}^2}\left[1-\mathrm{β}b{y}^3\right]dy}=\frac{∫\left[x_0-\mathrm{β}b{y}^4\right]e^{-\mathrm{β}a{y}^2}dy}{∫e^{-\mathrm{β}a{y}^2}dy}$

$\stackrel{¯}{x}=x_0-\mathrm{β}b\frac{\left(3/4{\mathrm{β}}^2{a}^2\right)\sqrt{\mathrm{π}/\mathrm{β}a}}{\sqrt{\mathrm{π}/\mathrm{β}a}}=x_0-\frac{3}{4}\frac{b}{\mathrm{β}{a}^2}=x_0-\frac{3}{4}\frac{b}{a^2}kT.$

07

part(d) Step 1:Given information

Lennard-Jones is positive

08

part(d) Step 2: Simplify

$\frac{d^{3}u}{d{x}^3}=u_0\left[(-12)(-13)(-14)x_0^{12}{x}^{-15}-2(-6)(-7)(-8)x_0^6{x}^{-9}\right]$

$\mathrm{Î\pm }=-\frac{3}{4}\frac{b}{a^2}\frac{k}{x_0}=-\frac{3}{4}\left(-\frac{252u_0}{x_0^3}\right){\left(\frac{x_0^2}{36u_0}\right)}^2\frac{k}{x_0}=\frac{7}{48}\frac{k}{u_0}=\frac{1.26×{10}^{-5}\mathrm{eV}/\mathrm{K}}{{\mathrm{u}}_0}.$

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