10. Appendix

630 Appendix B
The final step involves substituting in the expressions for q1min, q1max, q2min
and q2max from parts (a) and (b) and replacing k2 by (2m∗ /2 )Ek to obtain:
Ek 1/2 1
ELO
sinh 1
1/2
Ek
∝ NLO
Ùm
(NLO 1)
Ek
ELO
Ek
Ek 1/2 ELO
Ek
ELO
Ek
ELO
sinh 1
Ek ELO
ELO
1/2
(d) Readers are urged to substitute the parameters for GaAs into the above
expression to calculate Ùm.
Solution to Problem 5.6
(a) Starting with (5.72) we can write the drift velocity vd of charges q in an
applied electric field F and an applied magnetic field B as:
(m ∗ /Ù)vd q[F vd × B/c] (5.6a)
where m ∗ is the carrier effective mass and Ù is its scattering time. Let Ì
(qÙ/m ∗ ) represent the mobility of the charges. Then (5.6a) can be written as:
vd Ì[F vd × B/c] (5.6b)
Let N be the charge density, then the current density J is given by
J Nqvd qÌ[F vdxB/c] q [F (J × B/Nqc)] (5.6c)
For a magnetic field B applied along the z-direction (5.6c) can be written as:
Jx NqÌ[Fx (JyBz/Nqc)] (5.6d)
Jy NqÌ[Fy (JxBz/Nqc)] and (5.6e)
Jz NqÌFz
(5.6f)
The two equations (5.6d) and (5.6e) can be solved for the two unknowns:
Jx and Jy to give:
Jx NqÌ[Fx (ÌFyBz/c)]/[1 (ÌBz/c) 2 ] and (5.6g)
Jy NqÌ[Fy (ÌFxBz/c)]/[1 (ÌBz/c) 2 ] (5.6h)
To simplify the notation we introduce · NqÌ and ‚ ·Ì/c so that
(5.6g) and (5.6h) are rewritten as:
Jx [·Fx ‚FyBz)]/[1 (ÌBz/c) 2 ] and (5.6i)
Jy [·Fy ‚FyBz)]/[1 (ÌBz/c) 2 ] (5.6j)