____________________________________________________________________________________
1.9 Kinetic Molecular Theory Calculate the effective (rms) speeds of the He and Ne atoms in the He-Ne
gas laser tube at room temperature (300 K).
Solution
To find the root mean square velocity (vrms) of He atoms at T = 300 K:
The atomic mass of He is (from Periodic Table) Mat = 4.0 g/mol. Remember that 1 mole has a mass of Mat
grams. Then one He atom has a mass (m) in kg given by:
m=
M at
4.0 g/mol
=
× (0.001 kg/g ) = 6.642 × 10 − 27 kg
23
−1
N A 6.022 × 10 mol
From kinetic theory (visualized in Figure 1Q9-1),
1
3
2
m(v rms ) = kT
2
2
∴
v rms = 3
(
)
kT
1.381 × 10 -23 J K -1 (300 K )
= 1368 m/s
= 3
m
6.642 × 10 -27 kg
(
)
The root mean square velocity (vrms) of Ne atoms at T = 300 K can be found using the same method as
above, changing the atomic mass to that of Ne, Mat = 20.18 g/mol. After calculations, the mass of one Ne atom
is found to be 3.351 × 10-26 kg, and the root mean square velocity (vrms) of Ne is found to be vrms = 609 m/s.
Figure 1Q9-1: The gas molecule in the
container are in random motion.
Figure 1Q9-2: The He-Ne gas laser.
Author’s Note: Radiation emerging from the He-Ne laser tube (Figure 1Q9-2) is due to the Ne atoms emitting
light, all in phase with each other, as explained in Ch. 3. When a Ne atom happens to be moving towards the
observer, due to the Doppler Effect, the frequency of the laser light is higher. If a Ne atom happens to moving
away from the observer, the light frequency is lower. Thus, the random motions of the gas atoms cause the
emitted radiation not to be at a single frequency but over a range of frequencies due to the Doppler Effect.
1.31 Pb-Sn solder Consider the soldering of two copper components. When the solder melts, it wets both
metal surfaces. If the surfaces are not clean or have an oxide layer, the molten solder cannot wet the surfaces
and the soldering fails. Assume that soldering takes place at 250 ºC, and consider the diffusion of Sn atoms into
the copper (the Sn atom is smaller than the Pb atom and hence diffuses more easily).
a. The diffusion coefficient of Sn in Cu at two temperatures is D = 1.69 × 10-9 cm2 hr-1 at 400 ºC and D = 2.48
× 10-7 cm2 hr-1 at 650 ºC. Calculate the rms distance diffused by an Sn atom into the copper, assuming the
cooling process takes 10 seconds.
b. What should be the composition of the solder if it is to begin freezing at 250 ºC?
c. What are the components (phases) in this alloy at 200 ºC? What are the compositions of the phases
their relative weights in the alloy?
and
d. What is the microstructure of this alloy at 25 ºC? What are weight fractions of the α and β phases assuming
near equilibrium cooling?
Solution
a. Given information:
Temperatures: T1 = 400 °C + 273 = 673 K
T2 = 650 °C + 273 = 923 K
Diffusion coefficients: D1 = 1.69 × 10-9 cm2/hr = (1.69 × 10-9 cm2/hr)(0.01 m/cm)2 / (1 hr) × (3600 sec/hr)
D1 = 4.694 × 10-17 m2/s
D2 = 2.48 × 10-7 cm2/hr = (2.48 × 10-7 cm2/hr)(0.01 m/cm)2 / (1 hr) × (3600 sec/hr)
D2 = 6.889 × 10-15 m2/s
The diffusion coefficients D1 and D2 at T1 and T2 respectively are given by:
 qE 
D1 = Do exp − A 
 kT1 
 qE 
D2 = Do exp − A 
 kT2 
where EA is the activation energy in eV/atom and q = 1.6×10-19 J/eV (conversion factor from eV to J). Since
exp(− x)
= exp( y − x)
exp(− y )
we can take the ratio of the diffusion coefficients to express them in terms of EA (eV):
 qE 
Do exp − A 
D1
 kT1  = exp qE A − qE A  = exp q[T1 − T2 ]E A 
=



 kT
kT1T2
D2
 qE 


 2 kT1 
Do exp − A 
kT
2 

∴
D 
kT1T2 ln 1 
 D2 
EA =
q(T1 − T2 )
 4.694 × 10-17 m 2 /s 

1.381× 10 J K (673 K )(923 K ) ln
6.889 × 10-15 m 2 /s 

EA =
1.602 × 10-19 J/eV (673 K − 923 K )
(
∴
∴
(in eV)
- 23
(
-1
)
)
EA = 1.068 eV/atom
Now the diffusion coefficient Do can be found as follows:
(in eV)
 qE 
D1 = Do exp − A 
 kT1 
∴
Do =
D1
4.694 ×10-17 m 2 /s
= 4.638 × 10-9 m2/s
=
-19
 qE 
 1.602 ×10 J/eV (1.068 eV ) 

exp − A  exp −
1.381×10-23 J K -1 (673 K ) 

 kT1 
(
)
(
)
To check our value for Do, we can substitute it back into the equation for D2 and compare values:
(
)
 qE 
 1.602 ×10-19 J/eV (1.068 eV ) 

D2 = Do exp − A  = 4.638 × 10-9 m 2 /s exp −
1.381×10-23 J K -1 (923 K ) 

 kT2 
(
∴
)
(
)
D2 = 6.870 × 10-15 m2/s
This agrees with the given value of 6.889 × 10-15 m2/s for D2.
Now we must calculate the diffusion coefficient D3 at T3 = 250 °C + 273 = 523 K (temperature at which
soldering is taking place).
(
)
 qE 
 1.602 ×10-19 J/eV (1.068 eV ) 

D3 = Do exp − A  = 4.638 ×10-9 m 2 /s exp −
1.381×10-23 J K -1 (523 K ) 

 kT3 
(
∴
)
(
)
D3 = 2.391 × 10-19 m2/s
The rms distance diffused by the Sn atom in time t = 10 s (Lrms) is given by:
(
)
Lrms = 2 D3t = 2 2.391×10-19 m 2 /s (10 s ) = 2.19 × 10-9 m or 2 nm
b. From Figure 1Q31-1, 250 °C cuts the liquidus line approximately at 33 wt.% Sn composition (Co).
∴
Co = 0.33
(Sn)
c. For α-phase and liquid phase (L), the compositions as wt.% of Sn from Figure 1.69 or 1Q31-1 are:
Cα = 0.18
and
CL = 0.56
The weight fraction of α and L phases are:
C − Co 0.56 − 0.33
Wα = L
=
= 0.605 or 60 wt.% α-phase
CL − Cα 0.56 − 0.18
WL =
Co − Cα 0.33 − 0.18
=
= 0.395 or 39.5 wt.% liquid phase
CL − Cα 0.56 − 0.18
Figure 1.1Q31-1: The equilibrium phase diagram of the Pb-Sn alloy.
d. The microstructure is a primary α-phase and a eutectic solid (α + β) phase. There are two phases present, α
+β. (See Figure 1Q31-2)
Figure 1Q31-2: Microstructure of Pb-Sn at temperatures less than 183°C.
Assuming equilibrium concentrations have been reached:
C′α = 0.02
and
C′β = 1
The weight fraction of α in the whole alloy is then:
C ′ − Co 1 − 0.33
Wα′ = β
=
= 0.684 or 68.4 wt.% α-phase
Cβ′ − Cα′ 1 − 0.02
The weight fraction of β in the whole alloy is:
Wβ′ =
C′o − Cα 0.33 − 0.02
=
= 0.316 or 31.6 wt.% β-phase
Cβ′ − Cα′
1 − 0.02
------------------------------------------------------------------------------------------------------------------2.5 TCR and Matthiessen’s rule Determine the temperature coefficient of resistivity of pure iron and of
electrotechnical steel (Fe with 4% C), which are used in various electrical machinery, at two temperatures: 0 °C
and 500 °C. Comment on the similarities and differences in the resistivity versus temperature behavior shown in
Figure 2.39 for the two materials.
Solution
Figure 2Q5-1: Resistivity versus temperature for pure iron and 4% C steel.
The temperature coefficient of resistivity αo (TCR) is defined as follows:
αo =
Slope at To
1  dρ 
 =

ρ o  dT To
ρo
where the slope is dρ/dT at T = To and ρo is the resistivity at T = To.
To find the slope, we draw a tangent to the curve at T = To (To = 0 °C and then To = 500 °C) and obtain
∆ρ/∆T ≈ dρ/dT. One convenient way is to define ∆T = 400 °C and find ∆ρ on the tangent line and then
calculate ∆ρ/∆T.
Iron at 0 ºC:
Slopeo ≈ (0.23 × 10-6 Ω m − 0 Ω m) / (400 °C) = 5.75 × 10-10 Ω m °C -1
Since ρo ≈ 0.11 × 10-6 Ω m, αo = Slopeo/ρo ≈ 0.00523 °C -1
Fe + 4% C at 0 ºC:
Slopeo ≈ (0.57 × 10-6 Ω m − 0.4 × 10-6 Ω m) / (400 °C) = 4.25 × 10-10 Ω m °C -1
Since ρo ≈ 0.53 × 10-6 Ω m, αo = Slopeo/ρo ≈ 0.00802 °C -1
Iron at 500 ºC:
Slopeo ≈ (0.96 × 10-6 Ω m − 0.4 × 10-6 Ω m) / (400 °C) = 1.40 × 10-9 Ω m °C -1
Since ρo ≈ 0.57 × 10-6 Ω m, αo = Slopeo/ρo ≈ 0.00245 °C -1
Fe + 4% C at 500 ºC:
Slopeo ≈ (1.05 × 10-6 Ω m − 0.68 × 10-6 Ω m) / (400 °C) = 9.25 × 10-10 Ω m °C -1
Since ρo ≈ 0.85 × 10-6 Ω m, αo = Slopeo/ρo ≈ 0.00109 °C -1
*2.6 TCR of isomorphous alloys
a. Show that for an isomorphous alloy A%-B% (B% solute in A% solvent), the temperature coefficient of
resistivity αAB is given by
α AB ≈
α Aρ A
ρ AB
where ρAB is the resistivity of the alloy (AB) and ρA and αA are the resistivity and TCR of pure A. What are
the assumptions behind this equation?
b. Determine the composition of the Cu-Ni alloy that will have a TCR of 4×10-4 K-1, that is, a TCR that is an
order of magnitude less than that of Cu. Over the composition range of interest, the resistivity of the Cu-Ni
alloy can be calculated from ρCuNi ≈ ρCu + Ceff X (1−X), where Ceff, the effective Nordheim coefficient, is
about 1310 nΩ m.
Solution
a. By the Nordheim rule, the resistivity of the alloy is ρalloy = ρο + CX(1 − X). We can find the TCR of the
alloy from its definition
α alloy =
1
dρ alloy
ρ alloy dT
=
1
ρ alloy
d
[ρ o + CX (1 − X )]
dT
To obtain the desired equation, we must assume that C is temperature independent (i.e. the increase in the
resistivity depends on the lattice distortion induced by the impurity) so that d[CX(1 − X)]/dT = 0, enabling us to
substitute for dρo/dT using the definition of the TCR: αo =(dρo/dT)/ρo. Substituting into the above equation:
α alloy =
i.e.
1 dρ o
1
=
α o ρo
ρ alloy dT ρ alloy
α alloy ρ alloy = α o ρ o
or
α AB ρ AB = α A ρ A
Remember that all values for the alloy and pure substance must all be taken at the same temperature, or
the equation is invalid.
b. Assume room temperature T = 293 K. Using values for copper from Table 2.1 in Equation 2.19, ρCu = 17.1
nΩ m and αCu = 4.0 × 10-3 K-1, and from Table 2.3 the effective Nordheim coefficient of Ni dissolved in Cu is C
= 1310 nΩ m. We want to find the composition of the alloy such that αCuNi = 4 × 10-4 K-1. Then,
ρ alloy =
α Cu ρ Cu (0.0040 K −1 )(17.1 nΩ m)
=
= 171.0 nΩ m
0.0004 K −1
α alloy
Using Nordheim’s rule:
ρalloy = ρCu + CX(1 − X)
i.e.
171.0 nΩ m = 17.1 nΩ m + (1310 nΩ m)X(1 − X)
∴
X2─X+ 0.1175 = 0
solving the quadratic, we find X = 0.136
Thus the composition is 86.4% Cu-13.6% Ni. However, this value is in atomic percent as the Nordheim
coefficient is in atomic percent. Note that as Cu and Ni are very close in the Periodic Table this would also be
the weight percentage. Note: the quadratic will produce another value, namely X = 0.866. However, using this
number to obtain a composition of 13.6% Cu-86.4% Ni is incorrect because the values we used in calculations
corresponded to a solution of Ni dissolved in Cu, not vice-versa (i.e. Ni was taken to be the impurity).
*2.10 TCR and alloy resistivity Table 2.10 shows the resistivity and TCR (α) of Cu–Ni alloys. Plot TCR
versus 1/ρ, and obtain the best-fit line. What is your conclusion? Consider the Matthiessen rule, and explain
why the plot should be a straight line. What is the relationship between ρCu, αCu, ρCuNi, and αCuNi? Can this be
generalized?
NOTE: ppm-parts per million, i.e., 10-6.
Solution
The plot of temperature coefficient of resistivity TCR (α) versus 1/ρ, is as follows
TCR (α ) versus 1/resistivity (1/ρ)
-1
TCR, α (ppm ºC )
4500
4000
3500
3000
2500
2000
1500
1000
500
0
0
0.01
0.02
0.03
0.04
0.05
-1
0.06
1/Resistivity, 1/ρ (nΩ m )
Figure 2Q10-1: TCR (α) versus reciprocal of resistivity1/ρ
0.07
From Matthiessen’s rules, we have
ρ alloy = ρ matrix + ρ I = ρ o + ρ I
where ρo is the resistivity of the matrix, determined by scattering of electrons by thermal vibrations of crystal
atoms and ρI is the resistivity due to scattering of electrons from the impurities. Obviously, ρo is a function of
temperature, but ρI shows very little temperature dependence. From the definition of temperature coefficient of
resistivity,
αo =
and
1  δρ o 

 or
ρ o  δT 
α alloy =
δρo
= α o ρo
δT
δρo
α ρ
1  δρalloy  δ ( ρ o + ρ I )
1
 =

≈
= o o∝
ρ alloy  δT 
ρ alloyδT
ρ alloyδT ρ alloy ρ alloy
Clearly the TCR of the alloy is inversely proportional to the resistivity of the alloy. The higher the resistivity,
the smaller the TCR, which is evident from the plot.