Elements of Modern Physics
312
which comes out to be 2.1 × 10
–8
/kg mole (1.67 × 10
–6
/g mol in Gaussian units,
compared to the experimental value of 1.9 × 10
–6
/g mol).
Example 8
A ferromagnetic material with J = 3/2 and g = 2 has a transition temperature
T c = 120 K. Calculate the internal field near 0 K. What is the ratio of magnetization
at 300 K for B = 5 × 10
–3
T compared to that at 0 K?
From the expression for T c in Eq. (8.111), and
B int =
2
e g
N
J
m
λ
at 0 K,
one has
B int =
6
(
1)
c
mkT
e g J +
(8.169)
which is about 108 T. The ratio of magnetization at 300 K to that at 0 K, is
R =
0 (
1)
3 (
) 2
c
B J
e g
k T T
m
+
−
= 3.1 × 10
–5
(8.170)
which illustrates the fact that paramagnetic effects are, in general, much smaller
than ferromagnetic effects.
Example 9
When a photon is incident on a material with an energy gap ∆E, an electron in
the valence band may absorb this radiation and go to the conduction band if
hv > ∆E. The kinetic energy of the electron is given by
2
1
2
mv = h v – ∆E – (ε v – ε)
(8.171)
where ε is the initial energy of the valence electron, the maximum energy being
observed for ε = ε v . Thus, a rapid increase is observed in the absorptivity of
radiation as v increases through the value of v = ∆E/h. This property is used to
determine the energy gap (the experiments are usually done at low temperatures
to reduce thermal effects). Since ∆E ~ 1 eV for semiconductors, they are
essentially transparent to infrared radiation but absorb most of the radiation in
the optical region.
The excited electrons are de-excited either immediately, in general emitting
radiation (fluorescence) of a different frequency than that of the original photon,
or wander around in the crystal until they are trapped at the luminescent centers
312
which comes out to be 2.1 × 10
–8
/kg mole (1.67 × 10
–6
/g mol in Gaussian units,
compared to the experimental value of 1.9 × 10
–6
/g mol).
Example 8
A ferromagnetic material with J = 3/2 and g = 2 has a transition temperature
T c = 120 K. Calculate the internal field near 0 K. What is the ratio of magnetization
at 300 K for B = 5 × 10
–3
T compared to that at 0 K?
From the expression for T c in Eq. (8.111), and
B int =
2
e g
N
J
m
λ
at 0 K,
one has
B int =
6
(
1)
c
mkT
e g J +
(8.169)
which is about 108 T. The ratio of magnetization at 300 K to that at 0 K, is
R =
0 (
1)
3 (
) 2
c
B J
e g
k T T
m
+
−
= 3.1 × 10
–5
(8.170)
which illustrates the fact that paramagnetic effects are, in general, much smaller
than ferromagnetic effects.
Example 9
When a photon is incident on a material with an energy gap ∆E, an electron in
the valence band may absorb this radiation and go to the conduction band if
hv > ∆E. The kinetic energy of the electron is given by
2
1
2
mv = h v – ∆E – (ε v – ε)
(8.171)
where ε is the initial energy of the valence electron, the maximum energy being
observed for ε = ε v . Thus, a rapid increase is observed in the absorptivity of
radiation as v increases through the value of v = ∆E/h. This property is used to
determine the energy gap (the experiments are usually done at low temperatures
to reduce thermal effects). Since ∆E ~ 1 eV for semiconductors, they are
essentially transparent to infrared radiation but absorb most of the radiation in
the optical region.
The excited electrons are de-excited either immediately, in general emitting
radiation (fluorescence) of a different frequency than that of the original photon,
or wander around in the crystal until they are trapped at the luminescent centers