Answer to Question #164747 in Quantum Mechanics for Kumar

Question

Quantum mechanics is relevant, when the de Broglie wavelength of the particle is greater than the distance between particles. The purpose of this problem is to determine which systems will have to be treated quantum mechanically and which can be described classically.

a) Show that the typical de Broglie wavelength of a particle in an ideal gas in

equilibrium is 𝜆 =

ℎ

√3𝑚÷𝐾𝐵𝑇

b) Solids: The lattice spacing in a typical solid is d = 0.3 nm. Find the temperature below which the free electrons in a solid are quantum mechanical? (Hint: Refer the a) part of the question and treat free electrons as a gas and the lattice spacing as the typical distance between the electrons)

c) Gases: For what temperatures are the atoms in an ideal gas at pressure 𝑃 quantum mechanical? (Hint: Use the ideal gas law, to deduce the inter atomic distance)

Is Helium at atmospheric pressure quantum mechanical? What about Hydrogen atoms in outer space (interatomic distance is 1 cm and temperature is 3 K)?

Answer 1

a) The particles of an ideal gas that is not subjected to any potential only have kinetic energy. Each of the particles has an equal probability to move along the x, y and z directions of 3D space. So, it has three degrees of freedom. Therefore, the average kinetic energy of a particle is given by:

K=$\frac{3}{2}$k_BT

The kinetic energy can also be expressed as:

K=$\frac{1}{2}mv^2$ , where m is the mass of the particle, v is the root-mean-square velocity of the particle.

Equating the right-hand sides of the two equations above give us:

$\frac{3}{2}$k_BT = $\frac{1}{2}mv^2$ ⇒ 3k_BT=mv² ⇒ m²v² = 3mk_BT ⇒ mv =$\sqrt{3mK_BT}$

Now, the de Broglie wavelength of a particle is given by:

$\lambda = \frac{h}{mv}$, where h is Planck's constant

Inserting the value of mv from before, we get:

$\lambda = \frac{h}{\sqrt{3mk_BT}}$ λ=h√3mkBT

b) $\frac{h}{\sqrt{3mk_BT}}>d \implies T < \frac{h^2}{3mk_Bd^2}$

$T = \frac{(6.6\times{10}^{-34})^2}{3(9.1\times{10^{-31}})(1.4\times{10^{-23}})(3\times{10}^{-10})} = 1.3\times10^5 K$

c) Pd³ = k_BT $\implies d=(\frac{k_BT}{P})^{1/3}$

$T<\frac{h^2}{3mk_B}(\frac{P}{k_BT})^{2/3}\implies T<\frac{1}{k_B}(\frac{h^2}{3m})^{3/5}(P)^{2/5}$

For the helium m = m_He = 4m_p = 6.8 $\times$ 10^-27 kg

$T = \frac{1}{(1.4\times10^{-23})}(\frac{(6.6\times10^{-34})}{3(6.8\times10^{-27})})^{3/5}(1\times10^5)^{2/5}=2.8 K$

Answer: b) T < 1.3 $\times$ 10^-5 K; c) T < 2.8 K