For a system of bosons at room temperature, compute the average occupancy of a single-particle state and the probability of the state containing 0,1,2,3bosons, if the energy of the state is

(a) 0.001eVgreater than μ

(b) 0.01eVgreater than μ

(c) 0.1eVgreater than μ

(d) 1eVgreater than μ

Short Answer

Expert verified

Result is:

The energuy state is: (a).n¯=25.2

The energuy state is: (b).localid="1651000533351" n¯=2.10

The energuy state is: (c).localid="1651000556743" n¯=0.0208

The energuy state is: (d).localid="1651000583464" n¯=1.244×10-17

Step by step solution

01

Part(a) Step 1:Given information

We have been given that n¯=1e(ϵ-μ)/kT-1

02

Part(a) Step 2: Simplify

Here

P(n)=e-n(ϵ-μ)/kT1-e-(ϵ-μ)/kT

kT=8.62×10-5eV/K(298K)=0.02569eV

to find the average occupancy:

n¯=1e0.001eV/0.02569eV-1

n¯=25.2

if it contains n=1

P(1)=e-(1)(0.001eV)/0.02569eV1-e-0.001eV/0.02569eV

P(1)=0.03672

if it contains n=3

P(3)=e-(3)(0.001eV)/0.02569eV1-e-0.001eV/0.02569eV

P(3)=0.034

03

Part(b) Step 1:Given information

We have been given that ϵ-μ=0.01eV

04

Part(b) Step 2: Simplify

If it contain n=0

P(0)=e-(0)(0.01eV)/0.02569eV1-e-0.01eV/0.02569eV

P(0)=0.322

If it contains n=1

P(1)=e-(1)(0.01eV)/0.02569eV1-e-0.01eV/0.02569eV

P(1)=0.218

05

Part(c) Step 1: Given information

We have been given that ϵ-μ=0.1eV

06

Part(c) Step 2: Simplify

If it contains n=0

P(0)=e-(0)(0.1eV)/0.02569eV1-e-0.1eV/0.02569eV

P(0)=0.9796

If it contains n=1

P(1)=e-(1)(0.1eV)/0.02569eV1-e-0.1eV/0.02569eV

P(1)=0.020

07

Part(d) Step 1:Given information

We have been given that ϵ-μ=1eV

08

Part(d) Step 2: Simplify

If it contains n=0

P(0)=e-(0)(1eV)/0.02569eV1-e-1eV/0.02569eV

P(0)=1

If it contains n=1

P(1)=e-(1)(1eV)/0.02569eV1-e-1eV/0.02569eV

P(1)=1.244×10-17

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