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PracticeSol4

# PracticeSol4 - ME 083 Spring 2001 Structure and Properties...

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Unformatted text preview: ME 083 Spring 2001 Structure and Properties of Solids Practice Exam #4 April 25, 2001 E 4% /0 Problem 1: Semiconductors [((g‘ / — .- + + .- _ (a) (b) Figure 1: p-n junction diaode. Which of these two circuits will conduct electricity? Why? Of what engineering use is such a diode? <4) FQPMBO 6/er W 76 (aw;— Hm AfﬂoSS Yd/l/Cﬁbu /- Egg/‘gaw'w (b) An intrinsic semiconductor has a conductivity of 111 (Q——m)‘l at 10 ”C and 172 (SI—m)" at 17 "C. What is the band gape of this semicouductor? [0°C =£Q3L*/ ”2170;. £350 =a38667¢ __ £3. _; l / C9=(90‘(3 9256?— Q/' = 6 4421—9877 =82€§6 far-77: C "£34295 7:2. %(g2.= 03/5: ‘77—; / Eg=0886V (c) Suppose a piece of silicon is duped with phosphorous in a concentration of 1026 mi“. The intrinsic carrier concentration of silicon at room temperature is 2 x 10"5 m‘3. Calculate the resulting hole concentration at room temperature. 2. 5) 1 Z 26 / ‘ x _ A?— /x /0 P/5%sz>em~ae Liz—£6) 6 / 7:; A {/0 ) = Z} /0 -—-3 (c) GaAs (gallium arsenide) is a semiconducting material which can be made to convert elecn tricity to light with a very high efﬁciency. Actually, it is a semiconducting laser- Sketch the band structure you would expect for GaAs assuming it is an intrinsic semiconductor. Label all pertinent features. C‘EZJOW #3779763 014 Problem 2: Conductivity (a) A copper conductor is one mile long (1.62 x 105 em) and 0.3 cm in diameter. It carries a current of 96 A when the potential difference between its ends is 440 V (DC). Calculate the cross-sectional area of aluminum wire which would conduct the same current under the same conditions. The resistivity/6 of Aluminum is 6 x 10”“6 Q — cm _Z-e ~I‘ . _ 175-4? 4 x/ .Q- /6.2>r/o ll (b) Pure germanium has a. room temperature conductivity of 2 (Sl—m)‘1, and an equal number of negative and positive charge carriers with mobilities of 0.36 and 0.23 m2/ (V - sec), respectively. What fraction of the conductivity is due to electrons? 61: m/e/Mci‘ P/(f/ﬂf, ME? WNQQ/ 7., W W Wewp)§£‘ ﬂe C96 5 ”2/6/7697 (97: zW WP) 6%» 0,36 = ﬁ___________= O, 6 9 C973?” (0.36 + 0:23) / 3%76 6 039/ 09,:75 7/70/55 Problem 3: Mechanical Properties Despite advances in the development of metallic alloys, over 95% of all metals used today are iron and iron alloys. (a) A steel wire 10 inches in length with a diameter of 0.02 inches was pulled in tension with the following results: — yield point: 100 pounds — elongation at yield: 0.11 inch — maximum load: 120 pounds Determine the wires (i) elastic modulus, (ii) yield strength, and (iii) ultimate tensile strength. (L) (9:56 CC: 0'” 50,0” [email protected] .-.-.- M 7 /0 j Y 72: 0.03212 AME“ #Nwmgfo 75 M50,/ 6 '9‘ g: 6% g£?/[email protected]/WWW /o ,_ #2 my) (#X‘) (9073* W=O.38/[email protected] (b) At temperatures greater than approximately half the melting point, metallic alloys exhibit stress relaxation under load in a manner analogous to the low temperature viscoelastic behavior of linear polymers. At 0.7 Tm a steel beam is stretched (elongated) by a tensile stress of 50,000 psi, after which its length is held constant. If the time constant for stress relaxation is 60 days, calculate the magnitude of the stress you would expect after 30 days of exposure. IJ: (gage—(90.5? t /l 6:6004/ (90‘5‘30503/ 3004/ — goal Qamﬁowe/ '5’ = agozxgqmm 4 =30/321Zg/4Fw300L (c) For small—scale deformations, engineering stress and strain are good approximations for true stress and strain. Explain why. 3%: A}; (WW/Awavsﬂﬁa— 9/40 Wm ﬁamw (Gasser/7w ma; 57.1% f f €520 g ...
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PracticeSol4 - ME 083 Spring 2001 Structure and Properties...

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