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Unformatted text preview: 1.9 Kinetic Molecular Theory Calculate the effective (rrns) speeds of the He and Ne atoms in the
HeNe gas laser tube at room temperature (300 K). Solution To ﬁnd the root mean square velocity (12%) of He atoms at T = 300 K: The atomic mass of He is (from Periodic Table) Ma, = 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 a, _ 4.0 g/mol
NA 6.022x1023 mol_1 x (0.001 kg/g) = 6.642 x10"27 kg m: From kinetic theory (visualized in Figure 1Q91), lm(vms )2 = 2H" 2 2
23 ~1
Vms: BE: 31.381x10 JK 300K)=1368m/s +5
m 6.642 x10'27 kg The root mean square velocity (12%.) of Ne atoms at T = 300 K can be found using the same
method as above, changing the atomic mass to that of Ne, Ma = 20.18 g/mol. After calculations, the mass of one Ne atom is found to be 3.351 x 10'26 kg, and the root mean square velocity (vms) of Ne is found to
be v,,,,, = 609 m/s. +5 Figure 1Q91: The gas molecule in the Figure 1Q9_2: The He_Ne gas laser.
container are in random motion. Author ’5 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 ﬁequency 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 asingle frequency but over a range of frequencies due
to the Doppler Effect. 1.15 Thermal expansion a. If A. is the thermal expansion coefﬁcient, show that the thermal expansion coefﬁcient for an area is 2%.
Consider an aluminum square sheet of area 1 cm2. If the thermal expansion coefﬁcient of A1 at room temperature (25 °C) is about 24 x 10'6 K], at what temperature is the percentage change in the area
+1%? Solution a. Consider a rectangular area with sides x0 and yo. Then at temperature T 0,
A0 = xo yo and at temperature T , A = [x0 (1 + AAT)}[yO (1 + AAT)] = x0 y0 (1+ AMY +2
that is
A = x0 yo [1+ ZAAT + (AATY +2 We can now use that A0 = x0 y0 and neglect the term (AAT)2 because it is very small in comparison with
the linear term EAT (xi<<1) to obtain A = A0 (1 + MAT) = A0 (1 + 05A AT) +2
 So, the thermal expansion coefﬁcient for an area is
05A = 2/1
The area of the aluminum sheet at any temperature is given by
A=A0[1+ZA(T—TO)] +2 where A0 is the area at the reference temperature To. Solving for T we obtain, T=%+l_A_/A_9_—_1=250C+_1_ (1.01)—1 —————— : 2333 °C.
2 A. 2 24x10"6 °C”‘ +2 1.16 Thermal expansion of Si The expansion coefﬁcient of silicon over the temperature range
1201500 K is given by Okada and Tokumaru (1984) as 2 = 3.725 x 10‘6 [1 — 9‘5'33x10_3(T_124)] + 5.548 x10'wT Silicon linear expansion coeﬁ‘icient  where A is in K"1 (or °C '1) and T is in Kelvins. b. The change 6p in the density due to a change (ST in the temperature, from Example 1.8, is given by
(3/9 = —pOaV§T = —3po/16T Given density of Si as 2.329 g cm'3 at 20 0C, calculate the density at 1000 °C by using the full
expression, and by using the polynomials expansion of 2. What is your conclusiOn? NOTE: The exponential term is 5.88 x 10'3 NOT 3.725 x 10'3
NOTE: The example referred to is Example 1.8 NOT Example 1.5 Solution
b. The expansion coefﬁcient at 1000 °C (1273 K) will be 20273 K) = 3.725 x 10‘6 [1 — e‘533XIO‘3<1”3'124)] + 5.548 x 10400273) 2(1273 K) = 4.4269x10"6 K'l. +5
The density of Si at 293 K is 2.329 g cm'3, the density at 1000 °C (1273 K) is thus p=po + 5p = 1% +(—p.ayc5'1’)= p0  3/0016? = 2.329 g cm'3  3 (2.329 g cm'3)( 4.4269x10‘6 K'1)(1273 K — 293 K): 2.2987 g em'3. The density changes from 2.329 K to 2.2987 K which is very small change, therefore the density can be
roughly assumed as constant. EE271 Assignment Solutions Problem 2, ebooklet “Diffusion and Oxidation” Semiconductor Fabrication Consider a wafer of Si crystal which has uniform boron (B) doping of 1 X
1017 cm'3 in the bulk. Suppose that the Si wafer is exposed to a phosphorus (P) gas at 1250 °C for 10
minutes. During the diffusion process, the surface P concentration remains saturates at about 1 X 1021 cm" 3. Where is the junction from the surface? How long does it take to have the same junction depth from the
surface if the diffusion temperature is 1100 °C? State the assilmptions used in your calculations. The diffusion coefﬁcient of P in Si has D0 = 3 .85 cmzs'l, and Ed = 3.66 eV/atom. Solution Since we know, from the question, that during the diffusion process the surface P concentration
remains constant, this diffusion is from unlimited supply. As well as assuming that we have steady state
condition, we can now use Fick’s second law: 62C(x,t) _ 8C(x,t)
6x2 at Solving this differential equation with boundary conditions: D C (0, t) = C, at the surface at any time
C (x, t) = Co at t = 0 anywhere in the bulk
C (00, t) = Co at anytime at far end we have the following equation: where erf(z) is error function of z, and D is the diffusion coefﬁcient. Method using Table 1 Since we need to know the value of x when C(x, 6003) is the same concentration as boron, 17
1x10 0=1_erf[ x ] 1X100m'3, 1x1021—0 2m
1 1 17
erf x =1— X 021:0.9999
2th 1x10 From Table 1, we know that when erf(z) = 0.9999, 2 z 2.75. Thus, x
2.5: . At the diffusion temperature 1250 °C, the diffusion coefﬁcient of P atoms is: = 2.75 EE271 Assignment Solutions D = D0 exp(— RT 353119
= 3.85 X ex —————
8.3145  (1250+ 273) = 2.984 ><10‘12 cmzs'l
where Qdiﬁr is an activation energy in joule per mole,
Qdiﬂ= 3.66 [eV/atom] / q [C] ><NA = 353119 [J mol‘l]. Therefore, x 2J1} x=2.75x2JB7 = 2.75 X 2 2.984 X10_12 x 600
= 2.32 ><10'4 cm = 2.75 The junction is at 2.32 pm from the surface. EE271 Assignment Solutions Numerical Method The concentration proﬁle of P atoms at x at 10 min (600 sec) can be expressed as: C(x,t)—Co_1_erf( x ]
Cs—Ca 2J5? C(x,t) =l:1erf[2jﬁj:l(q —C0)—Ca
= Csl:1—ert{2 (.C,, = 0) C(x,600)=1><1021 x 1— erf[——i‘———]
2 2.984 x 1012 x 600 From this equation, the concentration proﬁle of P atoms at 10 min is graphically shown in Figure 1. 1E+21 1E+20 Phosphorus 1E+19 1E+18 1E+17 1E+16 1E+15 Concentration at 10 min, cm'3 1E+14
33 0.00 1.00 2.00 2' 3.00 4.00 5.00
Distance from the surface, um Figure 1 Concentration proﬁles of Phosphorus and Boron at 10 min. From this graph, we can see that the concentrations of phosphorus and boron atoms are equal at x = 2.33
pm, pn junction from the surface. We are now interested in how long it takes to have the same P concentration at x = 2.33 pm at the
diffusion temperature of 1100 °C. The P concentration at 1250 °C at 10 min at 2.33 pm, Clzsooc(2.33 um,
10 min), must be the same as the one 1100 °C at t’ at 2.33 m, Cnoooc(2.33 m, t’). Thus, EE271 Assignment Solutions C1250.C(2.33pm,10min) = C1100.C(2.33um,t') Cs[lw(2.%;J]=Cs[le«(2g71] Dt=D’t’
, D t =—t
D! 353119
2 XCX ————————
8.3145(1100+ 273) = 1.483 x10'13 cmzs'l
Therefore,
, Dt
t = —
DY
_ 2.984 ><10_12 x 600 1.483 ><10'13
= 12072.8 sec = 3.35 h ...
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This note was uploaded on 11/05/2011 for the course MSE 302 taught by Professor Norton during the Spring '11 term at Western Washington.
 Spring '11
 Norton

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