.9 Kinetic Mlecular Thery Calculate the effective (rms) speeds f the He and Ne atms in the He-Ne gas laser tube at rm temperature (300 K). Slutin T find the rt mean square velcity (v rms ) f He atms at T 300 K: The atmic mass f He is (frm Peridic Table) M at 4.0 g/ml. Remember that mle has a mass f M at grams. Then ne He atm has a mass (m) in kg given by: M m at N.0 g/ml 3 6.0 0 ml Frm kinetic thery (visualized in Figure Q9-), rms m v 3 ( ) kt 4 7-3 - kt (.38 0 J K )( 300 K) v rms 3 3 7 m ( 0.00 kg/g) 6.64 0 kg - ( 6.64 0 kg) 368 m/s The rt mean square velcity (v rms ) f Ne atms at T 300 K can be fund using the same methd as abve, changing the atmic mass t that f Ne, M at 0.8 g/ml. fter calculatins, the mass f ne Ne atm is fund t be 3.35 0-6 kg, and the rt mean square velcity (v rms ) f Ne is fund t be v rms 609 m/s. Figure Q9-: The gas mlecule in the cntainer are in randm mtin. Figure Q9-: The He-Ne gas laser. uthr s Nte: Radiatin emerging frm the He-Ne laser tube (Figure Q9-) is due t the Ne atms emitting light, all in phase with each ther, as explained in Ch. 3. When a Ne atm happens t be mving twards the bserver, due t the ppler Effect, the frequency f the laser light is higher. If a Ne atm happens t mving away frm the bserver, the light frequency is lwer. Thus, the randm mtins f the gas atms cause the emitted radiatin nt t be at a single frequency but ver a range f frequencies due t the ppler Effect..3 Pb-Sn slder Cnsider the sldering f tw cpper cmpnents. When the slder melts, it wets bth metal surfaces. If the surfaces are nt clean r have an xide layer, the mlten slder cannt wet the surfaces and the sldering fails. ssume that sldering takes place at 50 ºC, and cnsider the diffusin f Sn atms int the cpper (the Sn atm is smaller than the Pb atm and hence diffuses mre easily).
a. The diffusin cefficient f Sn in Cu at tw temperatures is.69 0-9 cm hr - at 400 ºC and.48 0-7 cm hr - at 650 ºC. Calculate the rms distance diffused by an Sn atm int the cpper, assuming the cling prcess takes 0 secnds. b. What shuld be the cmpsitin f the slder if it is t begin freezing at 50 ºC? c. What are the cmpnents (phases) in this at 00 ºC? What are the cmpsitins f the phases and their relative weights in the? d. What is the micrstructure f this at 5 ºC? What are weight fractins f the and β phases assuming near equilibrium cling? Slutin a. Given infrmatin: Temperatures: T 400 C + 73 673 K T 650 C + 73 93 K iffusin cefficients:.69 0-9 cm /hr (.69 0-9 cm /hr)(0.0 m/cm) / ( hr) (3600 sec/hr) 4.694 0-7 m /s.48 0-7 cm /hr (.48 0-7 cm /hr)(0.0 m/cm) / ( hr) (3600 sec/hr) 6.889 0-5 m /s The diffusin cefficients and at T and T respectively are given by: qe qe exp kt exp kt where E is the activatin energy in ev/atm and q.6 0-9 J/eV (cnversin factr frm ev t J). Since exp( x) exp( y x) exp( y) we can take the rati f the diffusin cefficients t express them in terms f E (ev): qe exp kt qe qe q[ T T ] E exp exp qe kt kt ktt exp kt ktt ln E (in ev) q ( T T ) -3-4.694 (.38 0 J K )( 673 K)( 93 K) ln 6.889-9 (.60 0 J/eV)( 673 K 93 K) -7 0 m /s -5 0 m /s E (in ev) E.068 ev/atm Nw the diffusin cefficient can be fund as fllws:
qe exp kt -7 4.694 0 m /s 4.638 0 ( )( ) -9 m /s -9 qe.60 0 J/eV.068 ev exp exp -3 - kt (.38 0 J K )( 673 K) T check ur value fr, we can substitute it back int the equatin fr and cmpare values: -9-9 ( ) (.60 0 J/eV)(.068 ev) 4.638 0 m /s exp qe exp kt K 6.870 0-5 m /s This agrees with the given value f 6.889 0-5 m /s fr. -3 - (.38 0 J K )( 93 ) Nw we must calculate the diffusin cefficient 3 at T 3 50 C + 73 53 K (temperature at which sldering is taking place). 3-9 -9 ( ) (.60 0 J/eV)(.068 ev) 4.638 0 m /s exp exp qe kt K 3 3.39 0-9 m /s -3 - (.38 0 J K )( 53 ) The rms distance diffused by the Sn atm in time t 0 s (L rms ) is given by: -9 (.39 0 m /s)( 0 s) L rms 3 t.9 0-9 m r nm b. Frm Figure Q3-, 50 C cuts the liquidus line apprximately at 33 wt.% Sn cmpsitin (C ). C 0.33 (Sn) c. Fr -phase and liquid phase (L), the cmpsitins as wt.% f Sn frm Figure.69 r Q3- are: C 0.8 and C L 0.56 The weight fractin f and L phases are: W C L C 0.56 0.33 0.605 r 60 wt.% -phase C L C 0.56 0.8 W L C C 0.33 0.8 0.395 r 39.5 wt.% liquid phase C L C 0.56 0.8
Figure.Q3-: The equilibrium phase diagram f the Pb-Sn. d. The micrstructure is a primary -phase and a eutectic slid ( + β) phase. There are tw phases present, +β. (See Figure Q3-) Figure Q3-: Micrstructure f Pb-Sn at temperatures less than 83 C. ssuming equilibrium cncentratins have been reached: C 0.0 and C β The weight fractin f in the whle is then: W C β C 0.33 C β C 0.0 The weight fractin f β in the whle is: W β C C C β C 0.33 0.0 0.0 0.684 r 68.4 wt.% -phase 0.36 r 3.6 wt.% β-phase -------------------------------------------------------------------------------------------------------------------.5 TCR and Matthiessen s rule etermine the temperature cefficient f resistivity f pure irn and f electrtechnical steel (Fe with 4% C), which are used in varius electrical machinery, at tw temperatures: 0 C and 500 C. Cmment n the similarities and differences in the resistivity versus temperature behavir shwn in Figure.39 fr the tw materials.
Slutin Figure Q5-: Resistivity versus temperature fr pure irn and 4% C steel. The temperature cefficient f resistivity (TCR) is defined as fllws: d dt T Slpe at T where the slpe is d/dt at T T and is the resistivity at T T. T find the slpe, we draw a tangent t the curve at T T (T 0 C and then T 500 C) and btain / T d/dt. One cnvenient way is t define T 400 C and find n the tangent line and then calculate / T. Irn at 0 ºC: Slpe (0.3 0-6 Ω m 0 Ω m) / (400 C) 5.75 0-0 Ω m C - Since 0. 0-6 Ω m, Slpe / 0.0053 C - Fe + 4% C at 0 ºC: Slpe (0.57 0-6 Ω m 0.4 0-6 Ω m) / (400 C) 4.5 0-0 Ω m C -
Since 0.53 0-6 Ω m, Slpe / 0.0080 C - Irn at 500 ºC: Slpe (0.96 0-6 Ω m 0.4 0-6 Ω m) / (400 C).40 0-9 Ω m C - Since 0.57 0-6 Ω m, Slpe / 0.0045 C - Fe + 4% C at 500 ºC: Slpe (.05 0-6 Ω m 0.68 0-6 Ω m) / (400 C) 9.5 0-0 Ω m C - Since 0.85 0-6 Ω m, Slpe / 0.0009 C - *.6 TCR f ismrphus s a. Shw that fr an ismrphus %-B% (B% slute in % slvent), the temperature cefficient f resistivity B is given by B B where B is the resistivity f the (B) and and are the resistivity and TCR f pure. What are the assumptins behind this equatin? b. etermine the cmpsitin f the Cu-Ni that will have a TCR f 4 0-4 K -, that is, a TCR that is an rder f magnitude less than that f Cu. Over the cmpsitin range f interest, the resistivity f the Cu-Ni can be calculated frm CuNi Cu + C eff X ( X), where C eff, the effective Nrdheim cefficient, is abut 30 nω m. Slutin a. By the Nrdheim rule, the resistivity f the is ο + CX( X). We can find the TCR f the frm its definitin d d dt dt [ + CX ( X )] T btain the desired equatin, we must assume that C is temperature independent (i.e. the increase in the resistivity depends n the lattice distrtin induced by the impurity) s that d[cx( X)]/dT 0, enabling us t substitute fr d /dt using the definitin f the TCR: (d /dt)/. Substituting int the abve equatin: d dt i.e. r B B Remember that all values fr the and pure substance must all be taken at the same temperature, r the equatin is invalid. b. ssume rm temperature T 93 K. Using values fr cpper frm Table. in Equatin.9, Cu 7. nω m and Cu 4.0 0-3 K -, and frm Table.3 the effective Nrdheim cefficient f Ni disslved in Cu is C 30 nω m. We want t find the cmpsitin f the such that CuNi 4 0-4 K -. Then,
Using Nrdheim s rule: (0.0040 K )(7.nΩ m) 0.0004 K Cu Cu Cu + CX( X) i.e. 7.0 nω m 7. nω m + (30 nω m)x( X) X X+ 0.75 0 slving the quadratic, we find X 0.36 7.0 nω m Thus the cmpsitin is 86.4% Cu-3.6% Ni. Hwever, this value is in atmic percent as the Nrdheim cefficient is in atmic percent. Nte that as Cu and Ni are very clse in the Peridic Table this wuld als be the weight percentage. Nte: the quadratic will prduce anther value, namely X 0.866. Hwever, using this number t btain a cmpsitin f 3.6% Cu-86.4% Ni is incrrect because the values we used in calculatins crrespnded t a slutin f Ni disslved in Cu, nt vice-versa (i.e. Ni was taken t be the impurity). *.0 TCR and resistivity Table.0 shws the resistivity and TCR () f Cu Ni s. Plt TCR versus /, and btain the best-fit line. What is yur cnclusin? Cnsider the Matthiessen rule, and explain why the plt shuld be a straight line. What is the relatinship between Cu, Cu, CuNi, and CuNi? Can this be generalized? NOTE: ppm-parts per millin, i.e., 0-6. Slutin The plt f temperature cefficient f resistivity TCR () versus /, is as fllws TCR, (ppm ºC - ) 4500 4000 3500 3000 500 000 500 000 500 0 TCR () versus /resistivity (/) 0 0.0 0.0 0.03 0.04 0.05 0.06 0.07 /Resistivity, / (nω m - ) Figure Q0-: TCR () versus reciprcal f resistivity/
Frm Matthiessen s rules, we have + + matrix I I where is the resistivity f the matrix, determined by scattering f electrns by thermal vibratins f crystal atms and I is the resistivity due t scattering f electrns frm the impurities. Obviusly, is a functin f temperature, but I shws very little temperature dependence. Frm the definitin f temperature cefficient f resistivity, and δ δt δ r δ δt I δt δt δt δ ( + ) δ Clearly the TCR f the is inversely prprtinal t the resistivity f the. The higher the resistivity, the smaller the TCR, which is evident frm the plt.