Question

In general, quantum mechanics is relevant when the de Broglie wavelength of the principle in question$\mathbf{}\mathbf{(}\mathit{h}\mathbf{/}\mathit{p}\mathbf{)}$ is greater than the characteristic Size of the system (d). in thermal equilibrium at (kelvin) Temperature$\mathit{T}$ the average kinetic energy of a particle is

$\frac{{\mathbf{p}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}{\mathit{k}}_{\mathbf{B}}\mathit{T}$

(Where ${\mathit{k}}_{\mathbf{B}}$is Boltzmann's constant), so the typical de Broglie wavelength is

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{h}}{\sqrt{\mathbf{3}\mathbf{m}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}}\mathbf{.}$

The purpose of this problem is to anticipate which systems will have to be treated quantum mechanically, and which can safely be described classically.

(a) Solids. The lattice spacing in a typical solid is around$\mathit{d}\mathbf{=}\mathbf{0}\mathbf{.3}\mathbf{}\mathbf{\text{nm}}$ . Find the temperature below which the free ${}^{\mathbf{18}}$electrons in a solid are quantum mechanical. Below what temperature are the nuclei in a solid quantum mechanical? (Use sodium as a typical case.) Moral: The free electrons in a solid are always quantum mechanical; the nuclei are almost never quantum mechanical. The same goes for liquids (for which the interatonic spacing is roughly the same), with the exception of helium below$\mathbf{4}\mathbf{}\mathbf{\text{K}}$ .

(b) Gases. For what temperatures are the atoms in an ideal gas at pressure quantum mechanical? Hint: Use the ideal gas law $\mathbf{(}\mathbf{P}\mathbf{V}\mathbf{=}\mathbf{N}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}\mathbf{)}$to deduce the interatomic spacing.

Short answer

(a) At a temperature below$3\text{K}$, we can treat the nuclei of sodium quantum mechanically.

(b) To treat the helium quantum mechanically we need it to be in a temperature less than$2.8\text{K}$.

Step-by-step solution

Step 1: Define the Schrodinger equation

• A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field.
• Its answer is related to a particle's probability density in space and time.

Determine the temperature to treat the sodium nuclei 

(a)

From the condition of treating a system quantum mechanically$(\lambda >d)$we can get the temperature.

$\frac{h}{\sqrt{3m{k}_{B}T}}>d=T<\frac{h^2}{3m{k}_B{d}^2}$ …(1)

To solve this problem we need to the mass of the electron$m_e=9.1\times {10}^{-31}\text{\hspace{0.17em}kg}$, the mass of the proton$m_p=1.7\times {10}^{-27}\text{\hspace{0.17em}kg}$, the value of Stefan-Boltzman constant, $k_B=1.4\times {10}^{-23}{\text{\hspace{0.17em}J}}^{\text{2}}/\text{kgK}$and the value of Plank constant$h=6.6\times {10}^{-34}\text{\hspace{0.17em}J.s}$. (n.b the mass of the nuclei is the number of protons and neutrons in that nuclei times the mass of the proton).

For free electrons:

Substitute all the value in equation (1)

$\begin{array}{c}T=\frac{{(6.6\times {10}^{-34}\text{\hspace{0.17em}Js})}^2}{3(9.1\times {10}^{-31}\text{\hspace{0.17em}kg})(1.4\times {10}^{-23}{\text{\hspace{0.17em}J}}^{\text{2}}/\text{kgK}){(3\times {10}^{-10}\text{\hspace{0.17em}m})}^2}\\ =1.3\times {10}^5\text{\hspace{0.17em}K}\end{array}$

So, at temperature less than$1.3\times {10}^5\text{\hspace{0.17em}K}$we can treat free electron in the solid quantum mechanically.

For Sodium nuclei: we have 23 particle in the nucleus, so

$\begin{array}{c}{m}_{\text{nuclei}}=23m_p\\ =23\times 1.7\times {10}^{-27}\text{\hspace{0.17em}kg}\\ m_{\text{nuclei}}=3.9\times {10}^{-26}\text{\hspace{0.17em}kg}\end{array}$.

Therefore,

$\begin{array}{c}T=\frac{{(6.6\times {10}^{-34}\text{\hspace{0.17em}Js})}^2}{3(3.9\times {10}^{-26}\text{\hspace{0.17em}kg})(1.4\times {10}^{-23}{\text{\hspace{0.17em}J}}^{\text{2}}/\text{kgK}){(3\times {10}^{-10}\text{\hspace{0.17em}m})}^2}\\ =3.0\text{\hspace{0.17em}K}\end{array}$

So, at a temperature below $3\text{K}$, we can treat the nuclei of sodium quantum mechanically.

Determine the temperatures are the atoms in an ideal gas at pressure

(b)

For one molecule (i.e.,$N=1,V=d^3$), and from the ideal gas equation the interatomic spacing is

$P{d}^3=k_{B}T\to d={\left(\frac{k_{B}T}{P}\right)}^{1/3}$

Using the condition$\lambda >d$and eq.(l) we can get the temperature.

$\frac{h}{\sqrt{3m{k}_{B}T}}>dT<\frac{h^2}{3m{k}_B{d}^2}$

Substitute$d$,

$T<\frac{h^2}{3m{k}_B}{\left(\frac{P}{k_{B}T}\right)}^{2/3}=T<\frac{1}{k_B}{\left(\frac{h^2}{3m}\right)}^{3/5}{\left(P\right)}^{2/5}$

For the helium

,$\begin{array}{c}m=m_{He}=4m_p\\ =4\times 1.7\times {10}^{-27}\text{\hspace{0.17em}kg}\\ =6.8\times {10}^{-27}\text{\hspace{0.17em}kg}\end{array}$

where$1\text{atm}=1\times {10}^5\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}}$,

$\begin{array}{c}T=\frac{1}{(1.4\times {10}^{-23}{\text{\hspace{0.17em}J}}^{\text{2}}/\text{kgK})}{\left(\frac{{(6.6\times {10}^{-34}\text{\hspace{0.17em}J.s})}^2}{3(6.8\times {10}^{27}\text{\hspace{0.17em}kg})}\right)}^{3/5}{(1\times {10}^5\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}})}^{2/5}\\ =2.8\text{K}\end{array}$

So to treat the helium quantum mechanically we need it to be in a temperature less than.

For hydrogen

,$\begin{array}{c}{m}_H=2m_p\\ =2\times 1.7\times {10}^{-27}\text{\hspace{0.17em}kg}\\ =3.4\times {10}^{-27}\text{\hspace{0.17em}kg}\end{array}$

with$d=1{\times }^2\text{m}$

$\begin{array}{c}T=\frac{{(6.6\times {10}^{-34}\text{\hspace{0.17em}J.s})}^2}{3(3.4\times {10}^{-27}\text{\hspace{0.17em}kg})(1.4\times {10}^{-23}{\text{\hspace{0.17em}J}}^{\text{2}}/\text{kgK}){(1\times {10}^{-2}\text{\hspace{0.17em}m})}^2}\\ =3.1\times {10}^{-14}\text{K}\end{array}$

So to see quantum mechanical behavior we need a temperature less than $3.1\times {10}^{-14}\text{K}$, therefore in the outer space the hydrogen show a classical behavior.