Example 9.4 investigated one criterion for quantum indistinguishability with reference to atmospheric nitrogen. Here we investigate the other, (a) Calculate the avenge separation between nitrogen molecules in the air. Assume a temperature of 300K a pressure of1atmand air that is 80% nitrogen. (b) Calculate the wave length of a typical nitrogen molecule in the air. (c) How does your result relate to Example 9.4?

Short Answer

Expert verified

a) The approximate separation between the nitrogen molecules is3.71×109m

b) The effective wavelength of the nitrogen molecules is2.76×1011m

c)4.12×107<<1means that the nitrogen molecules can be treated as a classical gas.

Step by step solution

01

The average distance, ideal gas law and the de Broglie's wavelength. 

The average distancebetween particles is proportional cube root of me volumeVper atom.

d(VN)1/3(1)

HereNis the number of particles.

A variation of the ideal gas law states that the product of the pressurePand volumeVis equal to the product of the number of particlesNBoltzmann's constantkBand temperatureT.PV=NkBT.(2)

The de Broglie's wavelength(λ)of an object is inversely proportional to the root of the massm.

Boltzmann's constantkBand temperatureTof the particle.

λ=h3mkBT

Here,nis Planck's constant.

The condition to see it an object can be treated classically is,

(λd)3<<1

02

The approximate separation between the nitrogen molecules  

(a)

Given data:

Convert of the molecular mass of nitrogen fromutokg.

m=28u=(28u)(1.66×1027kg1u)=4.65×1026kg

Rearrange equation (2) forVN.

VN=kBTP

SubstitutekBTPforVNin equation (1).

d(kBTP)1/3

Substitute1.38×1023J/KforkB,300Kfor1.01×105N/m2andfor

P.VN=(1.38×1023J/K)(300K)1.01×105N/m2=4.09×1026N/m2

Since the nitrogen is80%of the atmosphere, its partial pressure will be80%that of atmospheric pressure.

PN=80%(P)=(0.80)(1.01×105n/m2)=0.808×105N/m2

Substitute1.38×1023J/KforkB,300KforTand0.808×105N/m2for partial pressure due to nitrogen(P)in equation (3).

d=((1.38×1023J/K)(300K)0.808×105N/m2)13=3.71×109m

03

Find the effective wavelength of the nitrogen molecules.

(b)

Substitute6.63×1034j.s forh,4.65×1026kgform,1.38×1023J

I Kfor role="math" localid="1658385704993" kBand 300 k forT

λ=(6.63×1034j.s)3(4.65×1026kg)(1.38×1023J/K)(300K)=2.76×1011m

04

The nitrogen molecule can be treated as a classical gas. 

d(c)

The condition to see it an object can be treated classically is,

(λd)3<<1

Here,dis the separation that it has with other objects. If the statement is true, then the object can be treated as a classical gas. Otherwise, it's a quantum gas.

Calculation:

Substitute2.76×1011mforλand3.71×109mforin left hand side of in equality.

(λd)3<<1(λd)3=(2.76×1011m3.71×109m)3=4.12×107

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