Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
Question
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Chapter 38, Problem 36P
To determine

The mean free paths for the conduction electrons for Na, Au and Sn.

Expert Solution & Answer
Check Mark

Answer to Problem 36P

The mean free paths for the conduction electrons for Na, Au and Sn are 34.1nm,40.9nmand4.27nm respectively.

Explanation of Solution

Given:

The temperature is T=273K .

The resistivity and Fermi speed of Na is ρ1=4.2μΩcmand vF1=1.07×106m/s .

The resistivity and Fermi speed of Au is ρ2=2.04μΩcmand vF2=1.39×106m/s

The resistivity and Fermi speed of Sn is ρ3=10.6μΩcmand vF3=1.89×106m/s

Formula used:

The expression of mean free path is given by,

  λ=mevFne2ρ

Here, me is mass of electron, n is number of electrons per unit volume and e is the charge of electron.

Calculation:

Refer to the Table 38-1, the value of n for given elements is,

  nNa=26.5electrons/nm3 , nAu=59.0electrons/nm3 and nSn=148electrons/nm3 .

The mean free paths for the conduction electrons for Na is calculated as,

  λNa=( 9.109× 10 31 kg)( 1.07× 10 6 m/s )( 26.5 electrons/ nm 3 ) ( 1.602× 10 19 C )2( 4.2μΩcm)=( 9.109× 10 31 kg)( 1.07× 10 6 m/s )( 26.5× 10 22 electrons/ cm 3 ) ( 1.602× 10 19 C )2( 4.2μΩcm)=34.1nm

The mean free paths for the conduction electrons for Au is calculated as,

  λAu=( 9.109× 10 31 kg)( 1.39× 10 6 m/s )( 59.0 electrons/ nm 3 ) ( 1.602× 10 19 C )2( 2.04μΩcm)=( 9.109× 10 31 kg)( 1.39× 10 6 m/s )( 5.9× 10 22 electrons/ cm 3 ) ( 1.602× 10 19 C )2( 2.04μΩcm)=40.9nm

The mean free paths for the conduction electrons for Sn is calculated as,

  λSn=( 9.109× 10 31 kg)( 1.89× 10 6 m/s )( 148 electrons/ nm 3 ) ( 1.602× 10 19 C )2( 10.6μΩcm)=( 9.109× 10 31 kg)( 1.89× 10 6 m/s )( 14.8× 10 22 electrons/ cm 3 ) ( 1.602× 10 19 C )2( 10.6μΩcm)=4.27nm

Conclusion:

Therefore, the mean free paths for the conduction electrons for Na, Au and Sn are 34.1nm,40.9nmand4.27nm respectively.

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