91Ó°ÊÓ

We have given that a classical particle moving in a one-dimensional potential well, 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

The average position of equal summation over $x$for the positions weighted by their probabilities,

$\overline{x}=\underset{x}{∑}xP\left(x\right)$

The particle is inside one dimensional well, the probability:

$P\left(x\right)=\frac{e^{-\mathrm{β}u\left(x\right)}}{Z}$

By substituting it in above equation

$$\begin{gathered} \overline{x}=\underset{x}{∑}x\frac{e^{-\mathrm{β}u\left(x\right)}}{Z} \\ Z=\underset{x}{∑}{e}^{-\mathrm{β}u\left(x\right)} \\ \overline{x}=\frac{\underset{x}{∑}x{e}^{-\mathrm{β}u\left(x\right)}}{\underset{x}{∑}{e}^{-\mathrm{β}u\left(x\right)}} \end{gathered}$$

Multiply and Divide RHS by $∆x$

$\overline{x}=\frac{{\displaystyle \underset{x}{∑}}x{e}^{-\mathrm{β}u\left(x\right)}∆x}{\underset{x}{∑}{e}^{-\mathrm{β}u\left(x\right)}∆x}$

As $\underset{∆x→0}{\lim }$the summation becomes integral from $-∞to∞$

$\overline{x}=\frac{∫x{e}^{-\mathrm{β}u\left(x\right)}dx}{∫e^{-\mathrm{β}u\left(x\right)}dx}$

Hence proved.

We have given that a classical particle moving in a one-dimensional potential well , as shown. The particle is in thermal equilibrium with a reservoir at temperature , so the probabilities of its various states are determined by Boltzmann statistics

From figure we can say that the point $x_0$is local minimum,means slope is zero.

${\overline{)\frac{du}{dx}}}_{x_0}=0$

Therefore, Taylor series at point is

$$\begin{gathered} u\left(x\right)=u\left(x_0\right)+a{\left(x-x_0\right)}^2+b{\left(x-x_0\right)}^3 \\ a=\frac{1}{2}{\overline{)\frac{d^{2}u}{d{x}^2}}}_{x_0}b=\frac{1}{6}{\overline{)\frac{d^{3}u}{d{x}^3}}}_{x_0} \end{gathered}$$

approximation,

$u\left(x\right)≈u\left(x_0\right)+a{\left(x-x_0\right)}^2$

From part a

$$\begin{gathered} \overline{x}=\frac{∫x{e}^{-\mathrm{β}\left|u\left(x_0\right)+a{\left(x-x_0\right)}^2\right|dx}}{∫e^{-\mathrm{β}\left|u\left(x_0\right)+a{\left(x-x_0\right)}^2\right|dx}} \\ \overline{x}=\frac{∫x{e}^{-\mathrm{β}\left|a{\left(x-x_0\right)}^2\right|dx}}{∫e^{-\mathrm{β}\left|a{\left(x-x_0\right)}^2\right|dx}} \\ letx-x_0=y \\ dx=dy \\ \overline{x}=\frac{∫\left(y+x_0\right)e^{-\mathrm{β}a{y}^2}dy}{∫e^{-\mathrm{β}a{y}^2}dx} \\ \overline{x}=\frac{∫\left(y\right)e^{-\mathrm{β}a{y}^2}dy+∫\left(x_0\right)e^{-\mathrm{β}a{y}^2}dy}{∫e^{-\mathrm{β}a{y}^2}dx} \\ ∫\left(y\right)e^{-\mathrm{β}a{y}^2}dy=0functionisodd \\ \overline{x}=\frac{∫\left(x_0\right)e^{-\mathrm{β}a{y}^2}dy}{∫e^{-\mathrm{β}a{y}^2}dx} \\ \overline{x}=x_0 \end{gathered}$$

We have given that a classical particle moving in a one-dimensional potential well , as shown. The particle is in thermal equilibrium with a reservoir at temperature , so the probabilities of its various states are determined by Boltzmann statistics

If we keep cubic term in Taylor

$$\begin{gathered} \overline{x}=\frac{∫x{e}^{-\mathrm{β}|u\left(x_0\right)+a{\left(x-x_0\right)}^2+b{\left(x-x_0\right)}^3|dx}}{∫e^{-\mathrm{β}|u\left(x_0\right)+a{\left(x-x_0\right)}^2+b{\left(x-x_0\right)}^3|dx}} \\ letx-x_0=y \\ dx=dy \\ \overline{x}=\frac{∫\left(y+x_0\right)e^{-\mathrm{β}|u\left(x_0\right)+a{y}^2+b{y}^3|dy}}{∫e^{-\mathrm{β}|u\left(x_0\right)+a{y}^2+b{y}^3|dy}} \\ {e}^{-\mathrm{β}{y}^3}=1-\mathrm{β}b{y}^3 \\ \overline{x}=\frac{∫\left(x_0-\mathrm{β}b{y}^4\right)e^{-\mathrm{β}a{y}^2}dy}{∫e^{-\mathrm{β}a{y}^2}dy} \\ {∫}_{-∞}^{∞}{e}^{-\mathrm{β}a{y}^2}dy=\sqrt{\frac{\mathrm{π}}{\mathrm{β}a}} \\ {∫}_{-∞}^{∞}\mathrm{β}b{y}^4{e}^{-\mathrm{β}a{y}^2}dy=\frac{3b}{4\mathrm{β}{a}^2}\sqrt{\frac{\mathrm{π}}{\mathrm{β}a}} \\ \overline{x}=x_0-\frac{3bkT}{4a^2} \end{gathered}$$

We have given that a classical particle moving in a one-dimensional potential well , as shown. The particle is in thermal equilibrium with a reservoir at temperature , so the probabilities of its various states are determined by Boltzmann statistics

The Lennard Jones potential is given by$u\left(x\right)=u_0\left[{\left(\frac{x_0}{x}\right)}^{12}-2{\left(\frac{x_0}{x}\right)}^6\right]$

argon gas with

$u_0=0.010eV,x_0=3.9×{10}^{-10}m$

Numerically,

$$\begin{gathered} \mathrm{Î\pm }=\frac{1}{\overline{x}}\frac{∂\overline{x}}{∂T}=\frac{1}{\overline{x}}\frac{∂}{∂T}\left(x_0-\frac{3bkT}{4a^2}\right) \\ \mathrm{Î\pm }=\frac{1}{\overline{x}}\left(-\frac{3bk}{4a^2}\right) \\ \mathrm{Î\pm }≈-\frac{3bk}{4a^2{x}_0} \\ substitutewithaandb \\ \mathrm{Î\pm }=\frac{7k}{48u_0} \\ substitutewithgivenvaluesforargon \\ \mathrm{Î\pm }=\frac{7\left(8.62×{10}^{-5}eV/K\right)}{48\left(0.010eV\right)} \\ =1.257×{10}^{-3}{K}^{-1} \end{gathered}$$