Solid State Electronic Devices 7th Edition Streetman Solutions Manual

 

Chapter 2 Solutions Prob. 2.1 

(a&b) Sketch a vacuum tube device. Graph photocurrent I versus retarding voltage V for  several light intensities.

I

light intensity

Vo

V

 Note that Vo remains same for all intensities.  (c) Find retarding potential.

λ=2440 =0.244μm

Vo = h ν - Φ =

1.24eV  μm λ(μm)

=4.09eV -=

  1.24eV  μm - 4. 4.09 09eV eV = 5. 5.08 08eV eV - 4.09eV  1eV 0.244μm

Prob. 2.2 

Show third Bohr postulate equates to integer number n umber of DeBroglie waves fitting within circumference of a Bohr circular orbit.

4 o n2 rn = mq2

2

q2 mv2 an andd = and p  = mvr  4πo r 2 r 

2 2 2 2 π r 2 2 2 2 2   rn = 4 o n2 = n 2 4 o2 n = n 2  r n 2 = n2 2 mq mrB q mrn mv m v rn   m2v2rn 2 = n2 2   mvrn = n  p  = n is tthe he third third Bohr ohr ppos ostul tulate ate

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Prob. 2.3 

(a) Find generic equation for Lyman, Balmer, and Paschen series .  

mq4 hc ΔE = = λ 32π2o 2 n12 hc

 =

λ

mq4 ((nn 22 - n12 ) o 2n12 n 2 2 2 π2

32

2

=

mq4 32π2o2 n22 mq4 ((nn 22 - n12 )

2

8o2n12n22h2

8o 2 n12n 22 h2  hc 8εo2h3c n12n22   =  =   mq4 (n 22 - n12 ) mq4 n22 - n12 8(8.85  10-12 mF )2 (6.63 1034 Js)3 2.998  108 ms n12n 22   =  2 2 9.11  1100-31kg  (1.60  1100-19C)4 n 2 - n1 n12 n22 n12n22 8   = 9.1110 m  = 9.11Å  2 2 n22 - n12 n 2 - n1 n1 =1 for Ly Lyman, man, 2 for B Balmer, almer, and 3 fo forr Paschen (b) Plot wavelength versus n for Lyman, Balmer, and Paschen series. n 2 3 4 5

n^2 4 9 16 25

LYMAN SERIES n^2-1 n^2/(n^2-1) 3 1.33 8 1.13 15 1.07 24 1.04

LYMAN LIMIT n 3 4 5 6 7

n^2 9 16 25 36 49

BALMER SERIES n^2-4 4n^2/(n^2-4) 5 7.20 12 5.33 21 4.76 32 4.50 45 4.36

BALMER LLIIMIT

911*n^2/(n^2-1) 1215 1025 972 949

911Ǻ 911*4*n^2/(n^2-4) 6559 4859 4338 4100 3968

n 4 5 6 7 8 9 10

n^2 16 25 36 49 64 81 100

PASCHEN SERIES n^2-9 9*n^2/(n^2-9) 911*9*n^2/(n^2-9) 7 20.57 18741 16 14.06 12811 27 12.00 10932 40 11.03 10044 55 10.47 9541 72 10.13 9224 91 9.89 9010

PASCHEN LIMIT

8199Ǻ

3644Ǻ

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Prob. 2.4 

(a) Find Δ p x for   for Δx=1Ǻ.  Δpx  Δx =

h 6.6310-34J  s = = 5.03 10-25 kgsm    Δpx = -10 4   Δx 4π 10 m

h

4

(b ) Find Δt for Δ E=1eV.

ΔE  Δt = h

4

 Δt =

-15 h = 4.14 10 eV  s = 3.30 10-16s   4   ΔE 4π 1eV

Prob. 2.5 

 Find wavelength of 100eV and 12keV electrons. Comment on electron microscopes compared to visible light microscopes.

E= λ=

1 2

2 E m

mv2  v =

h

 p

=

h

mv

=

h

=

2 E m  

6.63 10-34J  s -31

2  9.1110 kg

  - 12

- 12

-19

1 2

 E = E  4.911100 J  m

For 100eV,  -1

-1

J 2 λ = E 2  4.911 10 0-19J 2  m = (100eV 1.602 10-19 eV ) 4.911 10 0-19J 2  m = 1.  1.2310-10m  = 1.23Å   1

1

For 12keV,  -1

-1

J 2 λ = E 2  4.911 10 0-19J 2  m = (1.2 104eeV V 1.602 10-19 eV )  4.911 10 0-19J 2  m = 1.12 10-11m =   0.112Å   1

1

The resolution on a visible microscope is dependent on the wavelength of the light which is around 5000Ǻ; so, the much smaller electron wavelengths provide much better resolution.  Prob. 2.6 

why? Assume 1-D in each Which of the following could NOT possibly be wave functions and why? case. (Here i= imaginary number, C is a normalization constant)

A) Ψ (x) (x) = C for for all x.  B) Ψ (x) = C for values of x between 2 and 8 cm, and Ψ (x) = 3.5 C ffor or values of x between 5 and 10 cm. Ψ (x) (x) is zero everywhere everywhere else. C) Ψ (x) = i C for x= 5 cm, and linearly goes down to zero at x= 2 and x = 10 cm from this peak value, and is zero for all other x. If any of these are valid wavefunctions, calculate C for those case(s). What potential energy for x

≤ 2 and x ≥ 10 is consistent with this?

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

A)  For a wavefunction (x (x)) ,

we know

Ρ=

  (x)(x)dx = 1 *

- 

Ρ=

  (x)(x)dx = c *

-

B)  For



2



0 c=0  (x) cannot be a wave function     c 0   

dx  Ρ= 

-

5  x  8 , (x) has two values, C and 3.5C. For c  0 , (x) is not a function 

and for c = 0 :



Ρ = * (x) (x)dx = 0  (

cannot be a wave function. 

-

iC  3  x-2 2  x  5   C)  (x)=    iC   x-10 5  x  10  5 

c2 c2 2 2 Ρ =   (x)(x)dx =   x-2 dx +   x-10 dx   - 2 9 5 25 *

5 10 c2 c2 (x-2)3  + (x-10)3  2 5 3×9 3×25   125  8c2 2  27 =c  + = 3  27 3×25 

=

8c2 Ρ=1  =1  c=0.612  (x) can be a wav wavee function   3 Since (x) = 0 for x  2  and x  10 , the potential energy should be infinite in these two

regions. 

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Prob. 2.7 A particle is described in 1D by b y a wavefunction: -2x +4x Ψ = Be  for x ≥0 and Ce  for x6 zero or non?

The energy barrier at x=0 is infinite; so, there is zero probability of finding an electron at x