Download ME280 Homework #12

ME280 Homework #12
PROBLEM SOLUTIONS
12.5
(a) Calculate the drift velocity of electrons in silicon at room temperature and when the magnitude of the
electric field is 500 V/m.
(b) Under these circumstances, how long does it take an electron to traverse a 25-mm (1-in.) length of crystal?
Solution
(a)
The drift velocity of electrons in Si may be determined using Equation 12.7. Since the room
temperature mobility of electrons is 0.14 m2/V-s (Table 12.3), and the electric field is 500 V/m (as stipulated in the
problem statement),
vd = e E
= (0.14 m2/V- s)(500 V/m) = 70 m/s

(b) The time, t, required to traverse a given length, l (= 25 mm), is just

t =

l
25  103 m
=
= 3.6  10-4 s
vd
70 m /s
12.31 Briefly explain why the ferroelectric behavior of BaTiO3 ceases above its ferroelectric Curie temperature.
Solution
The ferroelectric behavior of BaTiO3 ceases above its ferroelectric Curie temperature because the unit cell
transforms from tetragonal geometry to cubic; thus, the Ti4+ is situated at the center of the cubic unit cell, there is
no charge separation, and no net dipole moment.
12.D4 One of the procedures in the production of integrated circuits is the formation of a thin insulating layer of
SiO2 on the surface of chips (see Figure 12.26). This is accomplished by oxidizing the surface of the silicon by
subjecting it to an oxidizing atmosphere (i.e., gaseous oxygen or water vapor) at an elevated temperature. The rate
of growth of the oxide film is parabolic—that is, the thickness of the oxide layer (x) is a function of time (t)
according to the following equation:
x 2  Bt
(12.37)
Here the parameter B is dependent on both
temperature and the oxidizing atmosphere.
(a) For an atmosphere of O2 at a pressure of 1 atm, the temperature dependence of B (in units of μm2/h) is
as follows:
 1.24 eV 
B  800 exp 


kT 
(12.38a)
where k is Boltzmann’s constant (8.62 × 10–5 eV/atom) and T is in K. Calculate the time required to grow an oxide

layer (in an atmosphere of O2) that is 100 nm thick at both 700°C and 1000°C.
(b) In an atmosphere of H2O (1 atm pressure), the expression for B (again in units of μm 2/h) is
 0.70 eV 
B  215 exp 


kT 
(12.38b)
Now calculate the time required to grow an oxide layer that is 100 nm thick (in an atmosphere of H2O) at both

700°C and 1000°C, and compare these times with those computed in part (a).
Solution
(a) In this portion of the problem we are asked to determine the time required to grow a layer of SiO 2 that is
100 nm (i.e., 0.100 m) thick on the surface of a silicon chip at 1000C, in an atmosphere of O2 (oxygen pressure =
1 atm). Thus, using Equation 12.37, it is necessary to solve for the time t. However, before this is possible, we must
calculate the value of B from Equation 12.38a as follows:


 1.24 eV
1.24 eV
B  800 exp 

 = (800) exp 
-5

kT 

 (8.62  10 eV/atom - K)(1000 + 273 K)


= 0.00990 m2/h
Now, solving for t from Equation 12.37 using the above value for B and that x = 0.100 m, we have
x2
(0.100 m) 2
=
B
0.00990 m2 / h
t =
= 1.01 h

Repeating the computation for B at 700C:


1.24 eV
B = (800) exp 

-5

 (8.62  10 eV/atom - K)(700 + 273 K)

= 3.04  10-4 m2/h

And solving for the oxidation time as above
t =
(0.100 m) 2
3.04  10-4 m2 / h
= 32.9 h

(b) This part of the problem
asks for us to compute the heating times to form an oxide layer 100 nm thick at
the same two temperatures (1000C and 700C) when the atmosphere is water vapor (1 atm pressure). At 1000C,
the value of B is determined using Equation 12.38b, as follows:


 0.70 eV
0.70 eV
B  215 exp 

 = (215) exp 

kT 

 (8.62  10-5 eV/atom - K)(1000 + 273 K)

= 0.365 m2/h

And computation of the time t from the rearranged form of Equation 12.37, leads to
t =

And at 700C, the value of B is
x2
(0.100 m) 2
=
B
0.365 m2 / h
= 0.0274 h = 98.6 s


0.70 eV
B = (215) exp 
 = 0.0510 m2 / h
-5

 (8.62  10 eV/atom - K)(700 + 273 K)

Whereas the time required to grow the 100 nm oxide layer is

t =

x2
(0.100 m) 2
=
B
0.0510 m2 / h
= 0.196 h = 706 s
From the above computations, it is very apparent (1) that the 100 nm oxide layer forms more rapidly at
1000C (than at 700C) in both O2 and H2O gaseous atmospheres, and (2) that the oxide layer formation is more
rapid in water vapor than in oxygen.
16.3 An electrochemical cell is composed of pure copper and pure cadmium electrodes immersed in solutions of
their respective divalent ions. For a 6.5 × 10 –2 M concentration of Cd2+, the cadmium electrode is oxidized yielding
a cell potential of 0.775 V. Calculate the concentration of Cu 2+ ions if the temperature is 25°C.
Solution
The electrochemical reaction that occurs within this cell is just
Cd + Cu2+  Cd2+ + Cu
while V = 0.775 V and [Cd2+] = 6.5  10-2 M. Thus, Equation 16.20 is written in the form

V = (VCu  VCd ) 
0.0592
[Cd 2 ]
log
2
[Cu2 ]
This equation may be rewritten as


V  (VCu  VCd )
0.0296
 log
[Cd 2 ]
[Cu2 ]
Solving this expression for [Cu2+] gives


V  (VCu VPb) 

[Cu2+ ] = [Cd 2+ ] exp  (2.303)
0.0296




The standard potentials from Table 16.1 are VCu = +0.340 V and VCd = – 0.403 V. Therefore,


0.775 V  {0.340 V  (0.403 V)} 
[Cu2+ ] = (6.5  10-2 M ) exp  (2.303)


0.0296


= 0.784 M


16.7 A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to
experience a weight loss of 485 g due to corrosion. To what rate of corrosion, in both mpy and mm/yr, does this
correspond?
Solution
This problem asks for us to calculate the CPR in both mpy and mm/yr for a thick steel sheet of area 100
in.2 which experiences a weight loss of 485 g after one year. Employment of Equation 16.23 leads to
CPR(mm/yr) =
=
KW
A t
(87.6)(485 g)(10 3 mg / g)
(7.9 g /cm 3
)(100 in.2 ) (2.54 cm /in.)2 (24 h /day)(365 day / yr)(1 yr)
= 0.952 mm/yr

Also
CPR(mpy) =

(534)(485 g)(10 3 mg / g)
(7.9 g /cm 3)(100 in.2 ) (24 h /day)(365 day / yr)(1 yr)
= 37.4 mpy
16.12 For each form of corrosion, other than uniform, do the following:
(a) Describe why, where, and the conditions under which the corrosion occurs.
(b) Cite three measures that may be taken to prevent or control it.
For each of the forms of corrosion, the conditions under which it occurs, and measures that may be taken to
prevent or control it are outlined in Section 16.7.
16.13 Briefly explain why, for a small anode-to-cathode area ratio, the corrosion rate will be higher than for a
large ratio.
Solution
For a small anode-to-cathode area ratio, the corrosion rate will be higher than for a large ratio. The reason
for this is that for some given current flow associated with the corrosion reaction, for a small area ratio the current
density at the anode will be greater than for a large ratio. The corrosion rate is proportional to the current density (i)
according to Equation 16.24.
16.15 For each of the metals listed in the table, compute the Pilling–Bedworth ratio. Also, on the basis of this
value, specify whether or not you would expect the oxide scale that forms on the surface to be protective, and then
justify your decision. Density data for both the metal and its oxide are also tabulated.
Metal
Metal Density
(g/cm3)
Metal Oxide
Oxide Density
(g/cm3)
Mg
1.74
MgO
3.58
V
6.11
V2O5
3.36
Zn
7.13
ZnO
5.61
Solution
The general form of the equation used to calculate the Pilling-Bedworth ratios is Equation 16.32 (or
Equation 16.33). For magnesium, oxidation occurs by the reaction
Mg +
and therefore, from Equation 16.32
1
O  MgO
2 2

P  B ratio =
 =
AMgO  Mg
AMg MgO
(40.31 g / mol)(1.74 g /cm 3)
(24.31 g / mol)(3.58 g /cm 3)
= 0.81
Thus, this would probably be a nonprotective oxide film since the P-B ratio is less than unity; to be protective, this

ratio should be between one and two.
The oxidation reaction for V is just
2V +
5
O  V2O5
2 2
and the P-B ratio is (Equation 16.33)

P  B ratio =
AV O  V
2 5
(2) AV  V O
2 5

=
(181.88 g / mol)(6.11 g /cm 3)
(2)(50.94 g / mol)(3.36 g /cm 3)
= 3.25
Hence, the film would be nonprotective since the ratio does not lie between one and two.

Now for Zn, the reaction for its oxidation is analogous to that for Mg above. Therefore,
P  B ratio =
=
AZnO  Zn
AZn  ZnO
(81.39 g / mol)(7.13 g /cm 3)
(65.39 g / mol)(5.61 g /cm 3)
= 1.58
Thus, the ZnO film would probably be protective since the ratio is between one and two.
