HM_5_MSE_360_S15

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Dateof submission:

QUESTION3.

Thermaldiffusivity = 0.0015cm2/ s = 0.15×10-6m2/s

=

Whereα = thermal diffusivity and t = time taken.

Thus

0.4186=

Timet = ln0.4186 x

Timet = 10.75 hours

Time in hours

Temperature in Fahrenheit

QUESTION7

Thevoid through which the atom should jump is smaller than the atom’ssize and therefore more energy is needed to squeeze throughneighboring atoms. This is activation energy. Creating a vacancyrequires simply creating a space regardless of size. Thus activationenergy is higher than vacancy formation energy.

Inthe above figure, vacancies are first created using the vacancyformation energy after which the atoms can now jump into thevacancies as long as they have attained the minimum requiredactivation energy. The vacancies are represented by the dashedcircles while the jumping process is shown by the arrows.

QUESTION9

  1. Lattice diffusivity = 0.7 exp(-45/(4180 x T))

VaryingT from 400K to Tm 1234.78 K gave readings shown in the graph belowfor lattice diffusivity.

  1. Grain boundary diffusivity

=((0.5×10-9x 4.0863×10-10x/2)/π x d) x 0.09 exp(-21.5/(4.18 x T))

Whered is the grain size and T the temperature.

Valuesof the grain diffusivity for the different grain sizes are shown inthe graph below.

  1. Half of Tm = 617.39 K

Latticediffusivity = 0.7 exp (-45/ (4.18 x 617.39)) = 0.6879

0.6879= ((0.5×10-9x 4.0863×10-10x/2)/π x d) x 0.09 exp (-21.5/(4.18 x 617.39))

=whered is the required grain size

Solvingthe above equation: d = 5.9667 x 10-21m

T(K)

1/T

lnDeff (lattice)

lnDgrain(3µ)

lnDgrain(30µ)

lnDgrain(300µ)

lnDeff(30µ)

400

0.0025

-0.356701857

-0.342169957

-0.365195807

-0.388221658

-0.721897664

500

0.002

-0.356696475

-0.342169931

-0.365195782

-0.388221633

-0.721892257

600

0.00167

-0.356695886

-0.342189504

-0.365195765

-0.388221616

-0.721891651

700

0.00143

-0.356690323

-0.342189492

-0.365195752

-0.388221603

-0.721886075

800

0.00125

-0.3566884

-0.342189482

-0.365195743

-0.388221594

-0.721884143

900

0.00111

-0.356686905

-0.342189475

-0.365195736

-0.388221587

-0.721882641

1000

0.001

-0.356685709

-0.342189469

-0.36519573

-0.388221581

-0.721881439

1100

0.00091

-0.35668473

-0.342189465

-0.365195726

-0.388221577

-0.721880456

1200

0.00083

-0.356683915

-0.342189461

-0.365195722

-0.388221573

-0.721879637

1234.78

0.00081

-0.356683662

-0.34218946

-0.365195721

-0.388221571

-0.721879383

Table of naturallogs of the various diffusivities.

The MATLAB codeused is as below:

x=[0.00250.0020.001670.001430.001250.001110.0010.000910.000830.00081]

y=[-0.356701857-0.356696475-0.356695886-0.356690323-0.3566884-0.356686905-0.356685709-0.35668473-0.356683915-0.356683662]

y1=[-0.342169957-0.342169931-0.342189504-0.342189492-0.342189482-0.342189475-0.342189469-0.342189465-0.342189461-0.34218946]

y2=[-0.365195807-0.365195782-0.365195765-0.365195752-0.365195743-0.365195736-0.36519573-0.365195726-0.365195722-0.365195721]

y3=[-0.388221658-0.388221633-0.388221616-0.388221603-0.388221594-0.388221587-0.388221581-0.388221577-0.388221573-0.388221571]

y4=[-0.721892282-0.721892257-0.721891651-0.721886075-0.721884143-0.721882641-0.721881439-0.721880456-0.721879637-0.721879383]

plot(x,y,`color`,`blue`,`linewidth`,2)

holdon

plot(x,y1,`b`,x,y2,`b`,x,y3,`b`,x,y4,`b–`)

grid

text(0.0014,-0.3421,`grain3`)

text(0.002,-0.3653,`grain30`)

text(0.0014,-0.3884,`grain300`)

text(0.0025,-0.3567,`lattice`)

text(0.00167,-0.7219,`effective30`)

xlabel(`1/T`)

ylabel(`effectivediffusivity`)

title(`Diffusivityplots`)

text(0.001,-0.3883,`grain300`)

text(0.0025,-0.3836,`lattice`)

text(0.00167,-0.749,`effective`)

xlabel(`1/T`)

ylabel(`effectivediffusivity`)

QUESTION4

Layer= 10-7cm t = 100s Diffusivity = 10-12cm2/s

L=

=

=1 x 10-5

L2 =(1x 10-5)2

=1x 10-10

Surfaceconcentration = 1 x 10-10xx 1 x 10-5

= 1.7725 x 10-15cm-3

b.)At 105seconds:

L=

= 1 x 10-7

Thediffusion length is the same as the thickness of the depositedaluminum. The silicon chip is thicker than this, thus the solution isuseful as a semiconductor will be formed.

QUESTION6

  1. Diffusion over a distance is given by

X2= qiD t where qiis4 for 2-dimension diffusion D is the diffusion coefficient t istime

Makingtime the subject of the formula:

Timetaken = + + = 2.9225x 1011seconds.

Table of distance and concentration ratio

x

c/co

0.4690

0.9400

1.1705

0.4528

1.8493

0.1703

2.5846

0.0294

3.3121

0.0085

3.7376

0.0072

Plotof concentration ratio vs distance

x=[0.46901.17051.84932.58463.31213.7376]

y=[0.94000.45280.17030.02940.00850.0072]

plot(x,y,`k`)

grid

xlabel(`Distance`)

ylabel(`concentrationratio`)

title(`concentrationratio vs distance`)

Aboveis the MATLAB code used.

QUESTION12

ΔGsystem= XvΔHv- XvTΔSvvib– TΔSmix(Xv).

Solvefor the equilibrium value of Xv (do the algebra skipped in thenotes), by setting (∂ΔGsys/Xv)T,P,Xi,…=0.

Generally,

ΔGsystem= ΔHsystem– TΔS0system

Therefore:

ΔGsystem= Xv(ΔHv- TΔSvvib)- TΔSmix(Xv)

Takingthe partial derivative, we get:

Hv- TΔSvvib– TΔSmix= 0

+TΔSmix=ΔHv- TΔSvvib

(1+TΔSmixXv)=ΔHv- TΔSvvib

=ΔHv- TΔSvvib(1+TΔSmixXv)=0

(1+TΔSmixXv)=ΔHv- TΔSvvib

TΔSmixXv= ΔHv- TΔSvvib-1

Xv=

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