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2 and

(^2) n 2 2

4 π^2 2 2 2 o n (^) = n^ rn (^) = n

2

is the th

o o

n

Solid State Electronic Devices 7th Edition Streetman Solutions Manual

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Chapter 2 Solutions

Prob. 2.

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

Vo V

light in t ensity

Note that Vo remains same for all intensities.

(c) Find retarding potential.

λ=2440Å=0.244μm

V = hν - Φ = 1.24eV μm

=4.09eV

• =

1.24eV μm

• 4.09eV = 5.08eV - 4.09eV 1 eV o λ(μm) 0.244μm

Prob. 2.

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

4 n^2 r = q^2 mv^2 = and p = mvr n mq^2 4 n^2

4 π r^2 r  r

r = n o mq^2 mr 2 q^2 mr 2 mv^2 m^2 v^2 r

m^2 v^2 r 2 = n^2 mvrn = n

B n n

p = n ird Bohr postulate

(^2 32) π 2 2 2 2 o n 2

(^2) π 2 =^8

PASCHEN SERIES n n^ 2 n^ 2 - 9 9n^ 2 /(n^ 2 - 9 ) 9119n^ 2 /(n^ 2 - 9) 4 16 7 20. 57 18741 5 25 16 14. 06 12811 6 36 27 12. 00 10932 7 49 40 11. 03 10044 8 64 55 10. 47 9541 9 81 72 10. 13 9224 10 100 91 9. 89 9010 BALMER SERIES n n^ 2 n^ 2 - 4 4n^ 2 /(n^ 2 - 4 ) 9114*n^ 2 /(n^ 2 - 4) 3 9 5 7. 20 6559 4 16 12 5. 33 4859 5 25 21 4. 76 4338 6 36 32 4. 50 4100 7 49 45 4. 36 3968

o 1 2 1 2 1

2 1

Prob. 2.

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

h c mq^4 mq^4 ΔE = λ

= -

32 π^2 2 n 2

h c

mq^4 (n 2 - n 2 ) mq^4 (n 2 - n 2 )

λ 32 2 n 2 n 2 2 n 2 n 2 h^2 o 1 2 o 1 2 8 2 n 2 n 2 h^2 h c 8 ε 2 h^3 c n 2 n 2 = o^1 2 = o^1 mq^4 (n 2 - n 2 ) mq^4 n 2 - n 2 2 1 2 1 8(8.8510-12^ F^ )^2 (6.6310^34 Js)^3 2.99810^8 m^ n 2 n 2 = m^ s^1 9.1110-31kg (1.6010-19C)^4 n 2 - n 2

= 9.1110^8 m 1 n^22 n^2 n 2 - n 2

n 2 n 2 = 9.11Å 1 2 n 2 - n 2 2 1 2 1 n 1 =1 for Lyman, 2 for Balmer, and 3 for Paschen

(b) Plot wavelength versus n for Lyman, Balmer, and Paschen series. LYMAN SERIES n n^ 2 n^ 2 - 1 n^ 2 /(n^ 2 - 1 ) 911*n^ 2 /(n^ 2 - 1 ) 2 4 3 1. 33 1215 3 9 8 1. 13 1025 4 16 15 1. 07 972 5 25 24 1. 04 949 LYMAN LIMIT 911 Ǻ

PASCHEN LIMIT 8199 Ǻ

BALMER LIMIT 3644 Ǻ

2

1 1

eV

1 1 1 1

eV

1 1 1 1

Prob. 2. 4 (a) Find Δpx for Δx=1Ǻ.

• 34 Δp Δx =

h Δp =

h

6.6310 J s = 5.0310-^25 kgm

x 4  x 4 Δx 4 π 10 - 10 m s

(b) Find Δt for ΔE=1eV.

• 15

ΔE Δt = h Δt = h = 4.14 10 eV s = 3.3010-^16 s

4 4 ΔE 4 π 1eV

Prob. 2. 5 Find wavelength of 100eV and 12keV electrons. Comment on electron microscopes compared to visible light microscopes.

E = 1 mv^2 v =

2 E

m

• 34 λ =

h

h

h

6.6310 J s E

1 = E 4.9110 J m

• 2 - 2 p mv (^) 2 E m (^) 2 9.1110-^31 kg
• (^19 )

For 100eV,

λ = E

• 2

4.9110-19J^2 m = (100eV1.60210-^19 J^ )

• 2

4.9110-^19 J 2 m = 1.2310-^10 m = 1.23Å

For 12keV,

λ = E-^2 4.9110-19J^2 m = (1.210^4 eV1.60210-^19 J^ )-^2 4.9110-^19 J^2 m = 1.1210-^11 m = 0.112Å

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

A) Ψ (x) = C for all x.

B) Ψ (x) = C for values of x between 2 and 8 cm, and Ψ (x) = 3.5 C for values of x between 5 and 10 cm. Ψ (x) is zero 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?





• (^) 

A) For a wavefunction (x) , we know

 Ρ = *^ (x)(x)dx = 1

 Ρ = *^ (x)(x)dx = c^2 dx Ρ=

0 c = 0 (x) cannot be a wave function

• •  c^0

B) For 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 (xc) annot be a wave function.



C) (x)= 

iC x-2 3

2 x 5

iC x-10 5

5 x 10

(^5 2 10 ) c 2 c (^2) Ρ = (x)(x)dx = x-2 dx + x-10 dx

• 2 9

c^2 5 c^2

10

= (x-2)^3 3× 2 +^ (x-10)^3 3×25 5

27 125 ^ 8c^2 = c^2 + = 27 3×25 3

8c^2 Ρ = 1  =1 c=0. 3

(x) can be a wave function

Since (x) = 0 for x 2 and x 10 , the potential energy should be infinite in these two

regions.

2



Prob. 2. A particle is described in 1D by a wavefunction: Ψ = Be-2x^ for x ≥0 and Ce+4x^ for x<0, and B and C are real constants. Calculate B and C to make Ψ a valid wavefunction. Where is the particle most likely to be?

A valid wavefunction must be continuous, and normalized. For (0) = C = B

To normalize ,

 dx = 1

• 0  C^2 e8x^ dx + C^2 e-4xdx = 1 -  C^2

0 1 e8x^ + C^2 e-4x^1 8  C^2 C^2

+ = 1 C =

Prob. 2. The electron wavefunction is Ceikx^ between x=2 and 22 cm, and zero everywhere else. What is the value of C? What is the probability of finding the electron between x=0 and 4 cm? = Ceikx 22 *dx = C (^2) (20) = 1 2 4

C =

cm

20 2

Probability = ^

2 dx =

1 

2 =

Prob. 2. 9

Find the probability of finding an electron at x<0. Is the probability of finding an electron at x>0 zero or non-zero? Is the classical probability of finding an electron at x>6 zero or non?

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

x<0 (|ψ|^2 =0). However, it is possible for electrons to tunnel through the barrier at 5<x<6;

so, the probability of finding an electron at x>6 would be quantum mechanically greater

2 ^ A 2

2 ^ A

A e

(^2 28) ( 9. 11 10 - (^31) kg)

x z

 p 2  p 2  2

x z

  =   =

= 0





than zero (|ψ|^2 >0) and classical mechanically zero.

Prob. 2. 10

Find 4 p^2 2 p^2 7 mE for^ ( x ,^ y ,^ z ,^ t )^ A e j^ (10 x^3 y -4 t^ )^.

A*^ e-^ j(10x+3y-4t) 

ej(10x+3y-4t)dx

(^2) - j^ x x (^)  A e-^ j(10x+3y-4t)ej(10x+3y-4t)dx

• 

= 100

A*^ e-^ j(10x+3y-4t) 

ej(10x+3y-4t)dz

(^2) - j^ z z  A e-^ j(10x+3y-4t)ej(10x+3y-4t)dz

• 

A*^ e-^ j(10x+3y-4t)^  

j(10x+3y-4t)dt

E = -^

j t

A e-^ j(10x+3y-4t)ej(10x+3y-4t)dt

• 

= 4

4 p 2 +2 p 2 +7 mE = 400

2



L L





1

Prob. 2.

Find the uncertainty in position (Δx) and momentum (Δρ).

Ψ(x,t) =

sin πx e-2πjEt/ h^ and

L

• (^) dx = 1

L L (^0)

x = *^ x dx =

2 

x sin^2

x dx = 0.5L (from problem note)

0 L^0

L L

L

x^2 = *^ x dx =

2 

x^2 sin^2

x dx = 0.28L^2 (from problem note)

0 L^0 L

Δx = x^2 - x

2 = 0.28L^2 - (0.5L)^2 = 0.17L

p ^

h = 0. h

4 π Δx L

Prob. 2.

Calculate the first three energy levels for a 10Ǻ quantum well with infinite walls.

En = n^2 π^2  2 m L^2

(6.6310-34^ )^2

8 9.1110^31 (10^9 )^2

n^2 = 6.0310-20^ n^2

E = 6.0310-^20 J = 0.377eV E 2 = 4 0.377eV = 1.508eV E 3 = 9 0.377eV = 3.393eV

P r o b. 2. 1 3

Show schematic of atom with 1s^2 2s^2 2p^4 and atomic weight 21. Comment on its reactivity.

nucleus with 8 protons and 13 neutrons

2 electrons in 1s

2 electrons in 2s

This atom is chemically reactive because the outer 2p shell is not full. It will tend to try to add two electrons to that outer shell.

4 electrons in 2p

= proton = neturon = electron

Solid State Electronic Devices 7th Edition Streetman Solutions Manual

Full clear download (no error formatting) at:

https://testbanklive.com/download/solid-state-electronic-devices-7th-edition-

streetman-solutions-manual/