Question

Determine the total number (#/cm3) of energy states in silicon between Ec
and Ec + 2kT at (i) T = 300 K and (ii) T = 400 K.

Answer 1

Answer Density of allowed quantum state in the valence band is given by formula

g_v (E) = \frac {4 \pi {(2m_p)}^\frac {3}{2}}{h^3} \sqrt{E_v - E} (1)g

v

(E)=

h

3

4π(2m

p

)

2

3

E

v

−E

(1)

where h is the Planck's constant

Using (1) we have:

g_T (E) = \frac {4 \pi {(2m_p)}^\frac {3}{2}}{h^3} \int_{ E_v-kT }^ {E_v} \sqrt{E_v - E} dE (2)g

T

(E)=

h

3

4π(2m

p

)

2

3

∫

E

v

−kT

E

v

E

v

−E

dE(2)

After integrating (2) we got:

g_T (E) = \frac {4 \pi {(2m_p)}^\frac {3}{2}}{h^3} \frac {2}{3} {kT}^\frac {3}{2} (3)g

T

(E)=

h

3

4π(2m

p

)

2

3

3

2

kT

2

3

(3)

where k is the Boltzmann constant

In our case, mp=0.56m0 (m0=9.1×10-31 kg), T=300 K

Using (3) we got:

gT (E) = 4.08×1023 m-3

Answer:

4.08×1023 m-3

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Answer 2

The energy states in silicon between is $gT(E)=4.08*10^{23}m^{-3}$.

Explanation:

Density of allowed quantum state in the valence band is given,

$g_{v} (E)=\frac{4\pi (2m_{p})^{\frac{3}{2} } }{h^{3}} \sqrt{E_{v}-E}(1)$

where, h is the plan constant using (1) we have

gT(E)$=\frac{4\pi (2m_{p})^{\frac{3}{2} } }{h^{3}}$$g_{T} (E)=\frac{4\pi (2m_{p})^{\frac{3}{2} } }{h^{3}} \frac{2}{3} KT^\frac{3}{2} (3)$

$K$ is the Boltzmann constant

in our case, $m_{p}=0.56m_{0}$

$(m_{0}=9.1*10^{-31}kg),t=300k$

$gT(E)=4.08*10^{23}m^{-3}$

Therefore, the energy states in silicon is $gT(E)=4.08*10^{23}m^{-3}$.