A 1-dimensional crystal with 8 atoms (small, but illustrative) has a band structure given by: E (k) = -E0 cos (ka) (This is Ee (ke) when we are considering both e-'s and holes.) say Eo = 4eV = 4 (1.602 x 10-19 J) N = 8 (total number of atoms) a = 3Å = 3 x 10-10 m (atomic spacing) The allowed k's for cyclic boundary conditions are kn = 2π L n = 2π Na n n = −4, -3, -2, − 1, 0, 1, 2, 3, 4 Note: These are 9 (not 8) values of k, but -4 and 4 actually correspond to the same physical state (like the angles 0 and 2π) and should only be counted once. For example, we could count the n = 4 state and not the n = -4 state. So the k's would then be of the form: k_3 2π 8 3 x 10-10 m A) Calculate, tabulate and plot for the 8 allowed k's (n e E. (k.), V group, e (ke), meff,e(ke) (-3), etc., up to 4. = - -3, -2, ... 4): (our usual e quantities) B) Similarly, calculate, tabulate and plot: En (kn), Vgroup, h (kn), meff, h (kn) C) A full band would contain 2N = 16, e's and would have no net current. (remember, 2e¯ in each k-state, one spin up, one spin down.) Consider the case where 4e¯s are removed from the band (say, by thermal agitation), 2 from k4 and 2 from k3 (really ke, 4 and ke, 3). Indicate this situation on the Ee (ke) and Eh (kh) graphs. D) An electric field is applied in the (-x) direction such that the k-values of each e¯ are shifted by Ak +π = 12 Å (note, all k-values shift by equal amounts) i) Show what the e¯ configuration would be after the electric field is applied. Write down the expression for the e¯ current in the band, and evaluate the current. Remember I = (4) Σ (-e) v (k) (electron picture) (See pages 106-07 of notes.) {k} occupied ii) Show the hole configuration after the application of the field. Evaluate the hole expression for the current, and show that it gives numerically the same answer is in i). Note: holepicture 1 = Σ(+e) vn (kn) kh} "occupied" and the existence of an “occupied" hole (in the hole picture) is determined by the corresponding electron state which is empty (in the “real” electron picture).

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A 1-dimensional crystal with 8 atoms (small, but illustrative) has a band structure given by:
E (k) = -E0 cos (ka)
(This is Ee (ke) when we
are considering both e-'s and holes.)
say
Eo =
4eV = 4 (1.602 x 10-19 J) N = 8
(total number of atoms)
a =
3Å
=
3 x 10-10 m (atomic spacing)
The allowed k's for cyclic boundary conditions are
kn
=
2π
L
n =
2π
Na
n
n = −4, -3, -2, − 1, 0, 1, 2, 3, 4
Note: These are 9 (not 8) values of k, but -4 and 4 actually correspond to the same
physical state (like the angles 0 and 2π) and should only be counted once. For
example, we could count the n = 4 state and not the n = -4 state.
So the k's would then be of the form: k_3
2π
8 3 x 10-10 m
A) Calculate, tabulate and plot for the 8 allowed k's (n
e
E. (k.),
V
group, e (ke),
meff,e(ke)
(-3), etc., up to 4.
=
-
-3, -2, ... 4):
(our usual e quantities)
B) Similarly, calculate, tabulate and plot:
En (kn),
Vgroup, h
(kn),
meff, h (kn)
C) A full band would contain 2N = 16, e's and would have no net current. (remember, 2e¯
in each k-state, one spin up, one spin down.) Consider the case where 4e¯s are removed
from the band (say, by thermal agitation), 2 from k4 and 2 from k3 (really ke, 4 and ke, 3).
Indicate this situation on the Ee (ke) and Eh (kh) graphs.
Transcribed Image Text:A 1-dimensional crystal with 8 atoms (small, but illustrative) has a band structure given by: E (k) = -E0 cos (ka) (This is Ee (ke) when we are considering both e-'s and holes.) say Eo = 4eV = 4 (1.602 x 10-19 J) N = 8 (total number of atoms) a = 3Å = 3 x 10-10 m (atomic spacing) The allowed k's for cyclic boundary conditions are kn = 2π L n = 2π Na n n = −4, -3, -2, − 1, 0, 1, 2, 3, 4 Note: These are 9 (not 8) values of k, but -4 and 4 actually correspond to the same physical state (like the angles 0 and 2π) and should only be counted once. For example, we could count the n = 4 state and not the n = -4 state. So the k's would then be of the form: k_3 2π 8 3 x 10-10 m A) Calculate, tabulate and plot for the 8 allowed k's (n e E. (k.), V group, e (ke), meff,e(ke) (-3), etc., up to 4. = - -3, -2, ... 4): (our usual e quantities) B) Similarly, calculate, tabulate and plot: En (kn), Vgroup, h (kn), meff, h (kn) C) A full band would contain 2N = 16, e's and would have no net current. (remember, 2e¯ in each k-state, one spin up, one spin down.) Consider the case where 4e¯s are removed from the band (say, by thermal agitation), 2 from k4 and 2 from k3 (really ke, 4 and ke, 3). Indicate this situation on the Ee (ke) and Eh (kh) graphs.
D) An electric field is applied in the (-x) direction such that the k-values of each e¯ are shifted
by Ak
+π
=
12 Å
(note, all k-values shift by equal amounts)
i) Show what the e¯ configuration would be after the electric field is applied.
Write down the expression for the e¯ current in the band, and evaluate the current.
Remember I =
(4)
Σ (-e) v (k) (electron picture) (See pages 106-07 of notes.)
{k}
occupied
ii) Show the hole configuration after the application of the field. Evaluate the hole expression
for the current, and show that it gives numerically the same answer is in i).
Note: holepicture 1
=
Σ(+e) vn (kn)
kh}
"occupied"
and the existence of an “occupied" hole (in the hole picture) is determined by the
corresponding electron state which is empty (in the “real” electron picture).
Transcribed Image Text:D) An electric field is applied in the (-x) direction such that the k-values of each e¯ are shifted by Ak +π = 12 Å (note, all k-values shift by equal amounts) i) Show what the e¯ configuration would be after the electric field is applied. Write down the expression for the e¯ current in the band, and evaluate the current. Remember I = (4) Σ (-e) v (k) (electron picture) (See pages 106-07 of notes.) {k} occupied ii) Show the hole configuration after the application of the field. Evaluate the hole expression for the current, and show that it gives numerically the same answer is in i). Note: holepicture 1 = Σ(+e) vn (kn) kh} "occupied" and the existence of an “occupied" hole (in the hole picture) is determined by the corresponding electron state which is empty (in the “real” electron picture).
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