91影视

• 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.

(a)

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

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

To solve this problem we need to the mass of the electron$m_e=9.1脳{10}^{鈭�31}\text{鈥塳驳}$, the mass of the proton$m_p=1.7脳{10}^{鈭�27}\text{鈥塳驳}$, the value of Stefan-Boltzman constant, $k_B=1.4脳{10}^{鈭�23}{\text{鈥塉}}^{\text{2}}/\text{kgK}$and the value of Plank constant$h=6.6脳{10}^{鈭�34}\text{鈥塉.蝉}$. (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)

$\begin{array}{c}T=\frac{{(6.6脳{10}^{鈭�34}\text{鈥塉蝉})}^2}{3(9.1脳{10}^{鈭�31}\text{鈥塳驳})(1.4脳{10}^{鈭�23}{\text{鈥塉}}^{\text{2}}/\text{kgK}){(3脳{10}^{鈭�10}\text{鈥尘})}^2}\\ =1.3脳{10}^5\text{鈥块}\end{array}$

So, at temperature less than$1.3脳{10}^5\text{鈥块}$we can treat free electron in the solid quantum mechanically.

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

$\begin{array}{c}{m}_{\text{nuclei}}=23m_p\\ =23脳1.7脳{10}^{鈭�27}\text{鈥塳驳}\\ m_{\text{nuclei}}=3.9脳{10}^{鈭�26}\text{鈥塳驳}\end{array}$.

Therefore,

$\begin{array}{c}T=\frac{{(6.6脳{10}^{鈭�34}\text{鈥塉蝉})}^2}{3(3.9脳{10}^{鈭�26}\text{鈥塳驳})(1.4脳{10}^{鈭�23}{\text{鈥塉}}^{\text{2}}/\text{kgK}){(3脳{10}^{鈭�10}\text{鈥尘})}^2}\\ =3.0\text{鈥块}\end{array}$

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

(b)

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

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

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

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

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

,$\begin{array}{c}m=m_{He}=4m_p\\ =4脳1.7脳{10}^{鈭�27}\text{鈥塳驳}\\ =6.8脳{10}^{鈭�27}\text{鈥塳驳}\end{array}$

where$1\text{atm}=1脳{10}^5\text{鈥塏}/{\text{m}}^{\text{2}}$,

$\begin{array}{c}T=\frac{1}{(1.4脳{10}^{鈭�23}{\text{鈥塉}}^{\text{2}}/\text{kgK})}{\left(\frac{{(6.6脳{10}^{鈭�34}\text{鈥塉.蝉})}^2}{3(6.8脳{10}^{27}\text{鈥塳驳})}\right)}^{3/5}{(1脳{10}^5\text{鈥塏}/{\text{m}}^{\text{2}})}^{2/5}\\ =2.8\text{鈥�}\text{K}\end{array}$

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

For hydrogen

,$\begin{array}{c}{m}_H=2m_p\\ =2脳1.7脳{10}^{鈭�27}\text{鈥塳驳}\\ =3.4脳{10}^{鈭�27}\text{鈥塳驳}\end{array}$

with$d=1{脳}^2\text{m}$

$\begin{array}{c}T=\frac{{(6.6脳{10}^{鈭�34}\text{鈥塉.蝉})}^2}{3(3.4脳{10}^{鈭�27}\text{鈥塳驳})(1.4脳{10}^{鈭�23}{\text{鈥塉}}^{\text{2}}/\text{kgK}){(1脳{10}^{鈭�2}\text{鈥尘})}^2}\\ =3.1脳{10}^{鈭�14}\text{鈥�}\text{K}\end{array}$

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