Download Contact Potential Between Dissimilar Metals

Contact Potential Between Dissimilar Metals
Thomas R. Lemberger, Nov. 03
Contact Electric Field
A
Wire
B
Thermal Reservoir at τ.
When dissimilar metals, A and B, are connected by a wire, the internal constraint that
keeps electrons from flowing is lifted, and electrons will flow if doing so increases total entropy.
The electrostatic potential difference developed thereby is called the “contact potential”. In the
configuration shown, the system of two metals is in thermal contact with a bath. Hence, the
Helmholtz free energy, F, of the system is minimized by flow of electrons.
Problem: How do you know that equilibrium requires that the chemical potential of
electrons is the same in A as in B?
First, write down F as a function of number of electrons, Ne, transferred from A to B,
including the Coulomb energy in the capacitor, C, formed by the two metals:
F ( N e ) = FA ( N e ) + FB ( N e )
(eN e )2
+
.
2C
FA is the free energy of A when A is electrically neutral, FA,0, plus the change that occurs
when A gives up Ne electrons:
FA = FA, 0 +
∂FA, 0
∂N e, A
(− N e ) = FA,0 + µe, A (− N e ).
N A ,VA ,τ
µe,A is the change in free energy of A upon addition of one electron. µe,A is called the “internal”
chemical potential of electrons in A because it doesn’t include the “external” Coulomb energy.
Addition of many electrons requires us to include the Coulomb energy of transferred electrons
interacting with each other and with the positive charges left behind, i.e., the capacitor energy,
Q2/2C. A similar expression obtains for the free energy of metal B:
∂FB , 0
FB = FB , 0 +
∂N e, B
(+ N e ) = FB,0 + µe, B N e .
N B ,V B ,τ
.
Only a few electrons actually flow, so one derivative is enough in the Taylor expansion. Note
that the derivative that is the internal chemical potential is taken at fixed number of atoms, NA or
NB, as well as fixed volume and temperature.
Now we can minimize F(Ne) wrt Ne (at fixed τ, etc.) to find the equilibrium value of Ne:
F = FA, 0 − µ e, A N e + FB , 0 + µ e ,B N e
∂F
∂N e
= µ e,B
τ ,VA ,VB , N A , N B
2
(
eN e )
+
,
2C
e2 Ne
− µ e, A +
= 0.
C
We can rewrite, “e2Ne/C”, in terms of the electrostatic potential energy of electrons. The net
charge on the surface of metal A is eNe. If, for example, Ne is positive, then A is at a higher
voltage than B. The difference in voltage is: eN/C = φA – φB, where φ is the electrostatic
potential. The potential energy of electrons is –eφ, so we can write: -e2Ne/C = -e(φA – φB). Thus,
we can write the equilibrium condition as:
µ e , A − eϕ A = µ e , B − eϕ B .
As expected, in equilibrium the total chemical potential of electrons, internal chemical potential
plus external potential energy, is the same in metals A and B.