Chapter 4
4.1
⎪⎪⎭
⎫ ⎝
⎛-=kT
E N N n g
c i exp 2υ ⎪⎪⎭
⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=kT E T N N g O cO exp 3003
υ
where cO N and O N υ are the values at 300 K.
(b) Germanium
_______________________________________ 4.2
Plot
_______________________________________ 4.3
(a) ⎪⎪⎭
⎫
⎝⎛-=kT E N N n g c i exp 2υ (
)(
)(
)
3
19
19
2
113001004.1108.2105⎪⎭
⎫
⎝⎛⨯⨯=⨯T
()()⎥⎦
⎤⎢⎣⎡-⨯3000259.012.1exp T
()
3
382330010912.2105.2⎪⎭
⎫
⎝⎛⨯=⨯T
()()()()⎥⎦
⎤⎢⎣⎡-⨯T 0259.030012.1exp
By trial and error, 5.367≅T K
(b)
()
252
12
2105.2105⨯=⨯=i n
(
)
()()()()⎥⎦
⎤⎢⎣⎡-⎪⎭⎫
⎝⎛⨯=T T 0259.030012.1exp 30010912.23
38
By trial and error, 5.417≅T K
_______________________________________ 4.4
At 200=T K, ()⎪⎭
⎫
⎝⎛=3002000259.0kT
017267.0=eV
At 400=T K, ()⎪⎭
⎫
⎝⎛=3004000259.0kT
034533.0=eV
()()()()
172
22102
210025.31040.11070.7200400⨯=⨯⨯=
i
i n
n
⎥
⎦
⎤⎢⎣⎡-⎥⎦⎤⎢⎣⎡-⨯⎪⎭⎫
⎝⎛⎪
⎭⎫ ⎝⎛=017267.0exp 034533.0exp 3002003004003
3
g g E E
⎥⎦
⎤
⎢⎣⎡-=034533.0017267.0exp 8g g E E
()[]
9578.289139.57exp 810025.317-=⨯g E
or
()1714.38810025.3ln 9561.2817=⎪⎪⎭
⎫
⎝⎛⨯=g E or 318.1=g E eV
Now (
)
3
2
1030040010
70.7⎪⎭
⎫
⎝⎛=⨯o co N N υ
⎪⎭⎫ ⎝⎛-⨯034533.0318.1exp ()()
17
2110658.2370.210929.5-⨯=⨯o co N N υ so 371041.9⨯=o co N N υcm 6-
_______________________________________ 4.5
()()⎪⎭
⎫ ⎝⎛-=⎪⎭
⎫ ⎝⎛-⎪
⎭
⎫ ⎝⎛-=kT kT kT A n B n i i 20.0exp 90.0exp 10.1exp For 200=T K, 017267.0=kT eV For 300=T K, 0259.0=kT eV For 400=T K, 034533.0=kT eV
(a) For 200=T K, ()()6
10325.9017267.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i (b) For 300=T K, ()()4
1043.40259.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i (c) For 400=T K, ()()3
1005.3034533.020.0exp -⨯=⎪⎭
⎫ ⎝⎛-=A n B n i i _______________________________________ 4.6
(a) ()⎥⎦⎤
⎢⎣⎡---∝kT E E E E f g F c F c exp
()⎥⎦
⎤
⎢⎣⎡---∝kT E E E E c c exp
()⎥⎦⎤⎢⎣⎡--⨯kT E E F c exp Let x E E c =-
Then ⎪⎭⎫
⎝⎛-∝kT x x f g F c exp
To find the maximum value: ()⎪⎭
⎫
⎝⎛-∝-kT x x dx f g d F c exp 212/1
0exp 12/1=⎪⎭⎫ ⎝⎛-⋅-kT x x kT which yields
2212/12
/1kT
x kT x x =
⇒= The maximum value occurs at
2
kT
E E c +=
(b)
()()⎥⎦⎤
⎢⎣⎡---∝-kT E E E E f g F F exp 1υυ
()⎥⎦
⎤
⎢⎣⎡---∝kT E E E E υυexp
()⎥⎦⎤⎢⎣⎡--⨯kT E E F υexp Let x E E =-υ
Then ()⎪⎭
⎫ ⎝⎛-∝-kT x x f g F exp 1υ
To find the maximum value
()[]0exp 1=⎥⎦⎤
⎢
⎣⎡⎪⎭⎫ ⎝⎛-∝-kT x x dx d dx f g d F υ Same as part (a). Maximum occurs at
2
kT
x =
or
2
kT
E E -=υ
_______________________________________ 4.7
