10. Appendix

plus
Jz ·Fz
Solution to Problem 5.7 631
(5.6k)
If there are both electrons and holes we have to add their contributions to J.
Notice that · depends on the charge as q2 so ·n and ·p will both be positive.
Similarly (5.6i) to (5.6k) are valid for both electrons and holes. Finally, if we
assume that Bz is small then the B2 z terms can be neglected and the answer
for J is:
and
Jx (·n ·p)Fx (‚n ‚p)FyBz
Jy (·n ·p)Fy (‚n ‚p)FxBz
Jz (·n ·p)Fz
(b) In the Hall configuration Jy 0 in (5.6j) so that
Fy Fx[(‚n ‚p)Bz/(·n ·p)] and
Jx Fx[(·n ·p) 2 (‚n ‚p) 2 B 2 z ]/(·n ·p)
The Hall Coefficient is RH Fy/JxBz (‚n ‚p)/[(·n ·p) 2 (‚n ‚p) 2B2 z ].
Again, at low magnetic field, we can neglect the B2 z term so that
RH (‚n ‚p)/(·n ·p) 2 .
In terms of Nn and Np etc., the Hall Coefficient is equal to:
RH (1/qc)[Np (Ìn/Ìp) 2 Nn]/[(Ìn/Ìp)Nn Np] 2
Solution to Problem 5.7
For samples where the electron distribution is given by f (E) the ensemble
average of the current density is given by (5.88a) to(5.88c):
and
〈jx〉 ·Fx ÁBzFy
〈jy〉 ·Fy ÁBzFx
〈jz〉 〈Û0〉Fx
where F and B are the electric and magnetic fields, respectively; · and Á are
defined in (5.89a) and (5.89b). To obtain the Hall Coefficient RH we set 〈jy〉
0 so that Fy (Á/·)BzFx. Substituting back into RH Fy/〈Jx〉Bz we obtain:
RH (Á/· 2 )Fx/[·Fx ÁBzFy] (Á/· 2 )/[1 (ÁBz/·) 2 ].