30: A Microscopic View of Resistivity

INTRO:
Recall that the density J of current flowing through a material can be written in terms of microscopic properties of the material: j=nqv_d, where n is the density of current carriers, q is the charge of one current carrier, and v_d is the drift velocity of a current carrier. In a metal, the current carriers are electrons.
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PART A:
The drift velocity is the component of the current-carrier's velocity due to acceleration from the electric field in the conductor. This corresponds to the average speed of all of the current carriers in the conductor. The current carriers also have random thermal motions, but the randomness causes the velocities due to thermal motion to cancel when averaged over a large number of current carriers. If the electric field inside of the conductor has magnitude E, and the charge q is accelerated from rest for a time τ, what is the final speed v of the charge?
Express the speed in terms of E, q, τ, and the mass m of the charge.

SOLUTION:
The equation given by the intro is 

J = nqv_d (also eq. (30.13) from the book)

The book uses e as the charge, but q is an arbitrary charge, and more applicable to situations not concerning electrons. so q^* will be subbed into other equations that have e. (The asterick demonstrates a substituted variable)

Section 30.2 Creating a Current has a subsection titled: A Model of Conduction. This section is vital to this problem.

eq. (30.7) states that 

v_d = q^*τE/m
and that's actually the answer too

v = Eqτ/m

NOTE: 
At every collision, the electron's motion is randomized, bringing the drift velocity back to zero. If τ is the mean time between collisions, then the speed v that you just calculated is equal to the net drift velocity vd, for all current carriers. This is true, because the mean time between collisions is equal to the mean time since the last collision. (To see how this is possible requires looking at the actual distribution of times between collisions and how the different averages are calculated, but this type of analysis lies beyond the scope of this problem.)
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PART B:
The magnitude of the drift velocity is very small compared to the speed of random electron motion in a metal. The mean time between collisions can be calculated from the mean free path d, which is the average distance that an electron can travel before colliding with one of the metal nuclei. Using these variables, what is the mean time between collisions τ? Let E_F be the energy of the electrons.
Express the mean time between collisions in terms of E_F, d, and m.

SOLUTION:
so... how long, τ, does it take to go a distance, d, at a speed, v_F (the speed of an electron with kinetic energy equal to E_F)?

What equation relates energy velocity and mass? The kinetic energy equation!

K = ½mv²

E_F = ½mv_F²
→ v_F = SQRT{2E_F/m}

velocity is distance traveled over a period of time : v = d/t

therefore, distance traveled is equal to the velocity times time: d = v⋅t
Finally, that means that time elapsed is equal to distance traveled divided by velocity: t = d/v

∴ τ = d/v_F =d⋅ 1/SQRT{2E_F/m}

τ = d⋅SQRT{½m/E_F} = d⋅√{m/(2E_F)}

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PART C:
Recall that the conductivity σ of a substance is defined by the relation J=σE and that the resistivity ρ is the inverse of the conductivity: ρ=1/σ. Give an expression for ρ in terms of microscopic properties of a metal. Use q_e for the charge on the electron.
Express your answer in terms of E_F, m, q_e, d, E, and n.

SOLUTION:
so, if J=σE & ρ=1/σ, then J = E/ρ
eq(30.13) states that J = nev_d

∴ J = E/ρ = nev_d
→ρ = E/[nev_d]

if we plug in the value we got in part A for v, and the instructions statement that e = q_e ... 

ρ = E_F/[nq_e⋅(E_Fq_eτ/m)]

= m/[nq_e²⋅d⋅SQRT{m2E_F}]

If we then plug in the value we got in part B for τ... we should have everything in the proper variables. 

ρ = m/[nq_e²⋅τ]

=m/[nq_e²⋅d⋅SQRT{m/(2E_F)}]
=m⋅SQRT{2E_F/m}/[nq_e²⋅d]
=SQRT{2m²E_F/m}/[nq_e²⋅d]
⇒ ρ = SQRT{2mE_F} / [nq_e²⋅d]

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PART D:
Find the resistivity of gold at room temperature. Use the following information:

• free electron density of gold = 5.90×10²⁸ m⁻³
• Fermi energy of gold = 8.86×10⁻¹⁹ J
• mass of electron = 9.11×10⁻³¹ kg
• charge of an electron = −1.60×10−¹⁹ C
• mean free path of electron in gold = 3.45×10⁻⁸ m

Express your answer in ohm-meters to three significant figures.

SOLUTION:

Just plug these values into the equation we determined in the previous part... 

so what we were given was:

n = 5.90×10²⁸ m⁻³
E_F = 8.86×10⁻¹⁹ J
m = 9.11×10⁻³¹ kg
q_e = −1.60×10−¹⁹ C
d = 3.45×10⁻⁸ m

So ρ = SQRT{2mE_F} / [nq_e²⋅d] 

= SQRT{2(9.11×10⁻³¹ kg)(8.86×10⁻¹⁹ J)} / [(5.90×10²⁸ m⁻³)(−1.60×10−¹⁹ C)²⋅(3.45×10⁻⁸ m)]

⇒ ρ = 2.44×10^-8 Ωm
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