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 37P

(a)

To determine

The value of mean free path λi .

(a)

Expert Solution
Check Mark

Answer to Problem 37P

The value of mean free path λi is 66nm .

Explanation of Solution

Given:

The resistivity is ρ=1.0×108Ωm .

The Fermi energy from Table 38-1 is EF=7.03eV .

The value of ne(=N/V) for copper from Table 38-1 is ne=84.7electrons/nm3 .

Formula used:

The expression of resistivity from equation 38-13 is,

  ρ=mev avnee2λ=meuFnee2λi   ...... (1)

The expression for Fermi speed is,

  uF=2EFme   ...... (2)

Calculation:

From equation (1), the mean free path is,

  λi=meuFnee2ρ= 2 m e E F nee2ρ

Substitute values in above expression,

  λi= 2( 9.109× 10 31 kg )( 7.03eV )( 84.7 electrons/ nm 3 ) ( 1.602× 10 19 C )2( 1.0× 10 8 Ωm)= 2( 9.109× 10 31 kg )( 7.03eV )( 1.602× 10 19 J/ eV )( 84.7× 10 27 electrons/ m 3 ) ( 1.602× 10 19 C )2( 1.0× 10 8 Ωm)=66nm

Conclusion:

Therefore, the value of mean free path λi is 66nm .

(b)

To determine

The value of scattering cross section area.

(b)

Expert Solution
Check Mark

Answer to Problem 37P

The value of scattering cross section area is 1.8×104nm2 .

Explanation of Solution

Formula used:

The expression for mean free path from equation 38-16 is,

  λi=1neA   ...... (3)

Here, A is the scattering cross section area.

From equation (3), the scattering cross section area is,

  A=1neλi

Substitute values in above equation,

  A=1( 84.7 electrons/ nm 3 )( 66nm)=1.788×104nm21.8×104nm2

Conclusion:

Therefore, the value of scattering cross section area is 1.8×104nm2 .

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