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Solution Manual for Principles of Electronic Materials and Devices, 4th Edition Safa Kasap.pdf

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Solution Manual for Principles of Electronic Materials and Devices, 4th Edition Safa K Answers to "Why?" in the text Page 31: Oxygen has an atomic mass of 16 whereas it is 14 for nitrogen. The O2 molecule is therefore heavier than the N2 molecule. Thus, from 3( ) 2 2 1 2 1 mv  kT , th...

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Solution Manual for Principles of Electronic
Materials and Devices, 4th Edition Safa Kasap

,Answers to "Why?" in the text
Page 31:
Oxygen has an atomic mass of 16 whereas it is 14 for nitrogen. The O2 molecule is therefore heavier than
the N2 molecule. Thus, from 1
2 mv 2  3( 12 kT ) , the rms velocity of O2 molecules is smaller than that of N2
molecules.
Page 34, footnote 11
For small extensions, the difference between the engineering and instantaneous strains due to a
temperature change are the same. Historically, mechanical and civil engineers measured extension by
monitoring the change in length, L; and the instantaneous length L was not measured. It is not trivial to
measure both the instantaneous length and the extension simultaneously. However, since we know Lo and
measure L, the instantons length L = Lo + L. Is the difference important? Consider a sample of length
Lo that extends to a final length Ldue to a temperature change fromTotoT. Let  = (LLo) / Lo = L/Lo be
the engineering strain.
The engineering definition of strain and hence the thermal expansion coefficient is
L
Engineering strain =  T
Lo
so that thermal expansion from TotoT gives,
L
dL
T
L
L Lo  T dT 
Lo
  (T  To )     (T  To ) (1)
o o

where = L/Lo is the engineering strain as defined above
Physics definition of strain and hence the thermal expansion coefficient is
L
Instantaneous train =  T
L
so that thermal expansion from TotoT gives,
L T
 L
ln 1      (T  To )
dL
L L  T dT  ln     (T  To )
 Lo 
 (2)
o o

We can expand the ln(1 + ) term for small , so that Equation (2)essentially becomes Equation (1)

1.1 VirialtheoremThe Li atom has a nucleus with a +3e positive charge, which is surrounded by a full 1s
shell with two electrons, and a single valence electron in the outer 2s subshell. The atomic radius of the Li
atom is about 0.17 nm. Using the Virial theorem, and assuming that the valence electron sees the nuclear
+3e shielded by the two 1s electrons, that is, a net charge of +e, estimate the ionization energy of Li (the
energy required to free the 2s electron).Compare this value with the experimental value of 5.39 eV.
Suppose that the actual nuclear charge seen by the valence electron is not +e but a little higher, say
+1.25e, due to the imperfect shielding provided by the closed 1s shell. What would be the new ionization
energy? What is your conclusion?

,Solution
First we consider the case when the outermost valence electron can see a net charge of +e. From
Coulomb’s law we have the potential energy
Q1Q2 ( e)(e)
PE  
4πε0 r0 4πε0 r0

(1.6  10 19 C) 2
 = 1.354  1018 J or 8.46 eV
4 (8.85  10 12 Fm 1 )(0.17  10 9 m)

Virial theorem relates the overall energy, the average kinetic energy KE , and average potential energy
PE through the relations
1
E  PE  KE and KE   PE
2
Thus using Virial theorem, the total energy is
1
E PE  0.5  8.46eV = 4.23 eV
2
The ionization energy is therefore 4.23 eV.
Consider now the second case where the electronsees +1.25e due to imperfect shielding. Again
the CoulombicPE between +e and +1.25e will be
Q1Q2 (1.25e)(e)
PE  
4π 0 r0 4π 0 r0
1.25  (1.6  10 19 C) 2
 12 1 9
= 1.692  1018 J or10.58 eV
4π (85  10 Fm )(0.17  10 m)
The total energy is,
1
E PE  5.29 eV
2
The ionization energy, considering imperfect shielding, is 5.29 eV. This value is in closer
agreement with the experimental value. Hence the second assumption seems to be more realistic.

1.2 Virial theorem and the He atom In Example 1.1 we calculated the radius of the H-atom using the
Virial theorem. First consider the He+ atom, which as shown in Figure 1.75a, has one electron in the K-
sell orbiting the nucleus. Take the PE and the KE as zero when the electrons and the nucleus are infinitely
separated. The nucleus has a charge of +2e and there is one electron orbiting the nucleus at a radius r2.
Using the Virial theorem show that the energy of the He+ ion is
2e 2
E (He  )  () Energy of He+ ion [1.48]
4 o r2
Now consider the He-atom shown in Figure 1.75b. There are two electrons. Each electron interacts with
the nucleus (at a distance r1) and the other electron (at a distance 2r1). Using the Virial theorem show that
the energy of the He atom is
 7e 2 
E (He)  ()   Energy of He atom [1.49]
 8 o r1 

, The first ionization energy EI1 is defined as the energy required to remove one electron from the He
atom. The second ionization energy EI2 is the energy required to remove the second (last) electron from
He+. Both are shown in Figure 1.75 These have been measured and given as EI1 = 2372 kJ mole1 and
EI2= 5250 kJ mol1. Find the radii r1 and r2 for He and He+. Note that the first ionization energy provides
sufficient energy to take He to He+, that is, He He+ + e absorbs 2372 kJ mol1. How does your r1 value
compare with the often quoted He radius of 31 pm?

Figure 1.75: (a) A classical view of a He+ ion. There is one electron in the K-shell orbiting the nucleus that has a
charge +2e.(b) The He atom. There are two electrons in the K-shell. Due to their mutual repulsion, they orbit to
void each other.
Solution
Virial theorem relates the overall energy, the average kinetic energy KE , and average potential energy
PE through the relations
1 1 1
E  PE  KE ; KE   PE ; E  PE; KE   E (1)
2 2 2
Now, consider the PE of the electron in Figure 1.75a. The electron interacts with +2e of positive
charge, so that
(e)(2e) 2e 2
PE  
4 o r2 4 o r2
which means that the total energy (average) is
1 2e 2 e2
E (He  )  PE  ()  (2)
2 4 o r2 4 o r2
whichis the desired result.
Now consider Figure 1.75b. Assume that, at all times, the electrons avoid each other by staying in
opposite parts of the orbit they share. They are "diagonally"opposite to each other. The PE of this system
of 2 electrons one nucleus with +2e is

PE = PE of electron 1 (left) interacting with the nucleus (+2e), at a distance r1
+ PE of electron 2 (right) interacting with the nucleus (+2e), at a distance r1
+ PE of electron 1 (left) interacting with electron 2 (right) separated by 2r1
(e)(2e) (e)(2e) (e)(e)
 PE   
4 o r1 4 o r1 4 o (2r1 )
7e 2
 PE  
8 o r2

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