实验一：存在问题类型
实验一：存在问题类型 S D U T
a表1 Si粉衍射图指标化及晶胞参数d=2dsin0=aH?+K?+L?ka =0.15406nm/Nm峰号sin20mhkla(nm)Nm/Nid(nm)m3111110.543150.060340.31359280.543092202.670.160940.192013110.542973110.221393.670.163714160.543235.334000.321710.135815196.330.543163310.382140.1246186244220.543330.482390.1109179270.54329333(511)0.104510.543038324400.6437610.670.543090.0960193511.675310.704160.091800.5430710400.085870.543060.8047913.33620
表1 Si粉衍射图指标化及晶胞参数 峰号 sin 2θm Nm/N1 Nm hkl d(nm) a(nm) 1 2 3 4 5 6 7 8 9 10 0.06034 0.16094 0.22139 0.32171 0.38214 0.48239 0.54329 0.64376 0.70416 0.80479 1 2.67 3.67 5.33 6.33 8 9 10.67 11.67 13.33 3 8 11 16 19 24 27 32 35 40 111 220 311 400 331 422 333(511) 440 531 620 2d sin 2 2 2 a d H K L 0.15406 K nm 0.31359 0.19201 0.16371 0.13581 0.12461 0.11091 0.10451 0.09601 0.09180 0.08587 0.54315 0.54309 0.54297 0.54323 0.54316 0.54333 0.54303 0.54309 0.54307 S D U T 0.54306
acal=ao+b.cos'edncos2日ma cal-1(a-a cal-1)2acal-2(a-acal-2)2a.nm0.9396601.068601E-100.5431370.5431371.068601E-100.8390600.5431590.5431334.512247E-091.695238E-090.5431300.7786100.5431563.382909E-082.494089E-080.6782900.5431517.034899E-090.5431251.207236E-080.6178600.5431488.053894E-110.5431221.223205E-090.5431190. 5176103.608949E-080.5431170.5431434.664405E-080.4567100.5431411.164038E-080.5431156.706066E-090.3562400.5431361.973603E-090.5431103.394897E-100.2958400.5431333.587683E-090.5431071.149025E-090.1952100.5431284.516680E-090.5431021.697955E-09a0=0.5431191.033715E-07a0=0.5430939.657515E-08
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点BierosoftExcel-新建ieresoftErcel T作表ls键入需要帮助的可题国文件）编摄）：初图V）插入格式(）THT)数据甲窗口帮助(0).6100%Oe教B卫非三国号%年国·9·Ai图片5-CDEFBGHA4LMNT1singNSONsin2ed.racos28ma_cal-1(a-a cal-1)2α_cal-2(a-a_cal-2)2a,nm23.00.060340.2456421.7320510.3135860.5431480.9396600.5431371.068601E-100.5431371.068601E-1038.00.160940.4011732.8284270.1920120.5430910.8390600.5431594.512247E-090.5431331.695238E09430.221390.47052111.03.3166250.1637120.5429720.7786100.5431563.382909E-080.5431302.494089E-0850.3217116.00.1358090.67829040.5671954.0000000.5432350.5431517.034899E-090.5431251.207236E-08619.00.382140.6181754.3588990.1246090.5431570.6178600.5431488.053894E-110.5431221.223205E-090·543119700.482390.69454324. 04.8989790.1109070.5433330.5176100.5431433.608949E080.5431174.664405E088t0.543290.73708227.05.1961520.1045070.5430330.5431411.164038E-080.4567100.5431156.706066E090090.80234732.0H0.5431360.643765.6568540.0960060.5430910.3562401.973603E-090.5431103.394897E-10100.7041635.00.8391425.9160800.0917960.5430730.2958400.5431333.587683E-090.5431071.149025E-0910110.804790.89710140.00.0858650.5430610.1952100.5431284.516680E-090.5431021.697955E-096.32455512a0=0.5431191.033715E-07a0=0.5430939.657515E-0813014O15b0.000047a_cal=ao+b.cos'e0.5440160.5438170.5436180.543419ao=5.43088A±3.5·10-5A0.543220210.5430USBureauofStandardsStandardRefe220.5428ao=5.43054A±1.1-10A(2ppm.F230.542624A±6.7.10*A(12ppm).Intermational0.5424Equationy=a+b'x250.54221950.Adj R-Square0.111122605420-ValueStandardError国WSheetiSheet2(Sheet3绘图自图形、口O国·A#数字就绪22.46WV开始作业安排doe.XAo1OOOCdTHBOL在线教实验一Originro8实验一答案微信2Micresoft2023/3/29
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a= a +b.cos?00.54400.54380.54360.54340.54320.54300.54280.5426wu'e0.5424Equationy=a+b*x0.5422Adj.R-Square0.111120.5420ValueStandardErrorB9.14423E-5Intercept5.430927E-10.5418BSlope4.711493E-51.49045E-40.54160.54140.54120.54100.10.20.30.40.50.60.70.80.91.0cose
S D U T
2、物相鉴定一一定性分析表2XRD谱图实验数据%%d2 0%2 02 0dCpsCpsdCps7.712160133.234.6023735766.5757913.4542.590201.8251049.9293807.97.9464.3060720. 612.5356935.3737852.19322141.75113174036.1446292. 664113.33.7974723.4072.4870436.0851.5905457.93365417.33.2894327.086456294.72.3093938.96913.61.5100961.341836163647. 62.8801931.0252.1779341.426163333.91.4534164.01772481810047.8364038.455811.62.6576933.6971.899951.4362464.868蓝色：β-Si3N4黑色：α-Si3N4
表2 XRD谱图实验数据 d 2θ Cps % d 2θ Cps % d 2θ Cps % 6.57579 13.454 1601 33.2 2.59020 34.602 373 7.7 1.82510 49.929 576 12 4.30607 20.61 380 7.9 2.53569 35.37 378 7.9 1.75113 52.193 2214 46 3.79747 23.407 1740 36.1 2.48704 36.085 4462 92.6 1.59054 57.933 641 13.3 3.28943 27.086 4562 94.7 2.30939 38.969 654 13.6 1.51009 61.341 836 17.3 2.88019 31.025 364 7.6 2.17793 41.426 1633 33.9 1.45341 64.01 772 16 2.65769 33.697 4818 100 1.89995 47.836 403 8.4 1.43624 64.868 558 11.6 2、物相鉴定——定性分析 蓝色： β-Si3N4 黑色： α-Si3N4 S D U T
ββ(101)(200)B(120)() sβββ(100)(301)(110)β(201)aβ(210)βββ(002)Baβ(310)(221X320)βα (102)(111)(101)(201)(220)(410)40102030506070
10 20 30 40 50 60 70 I n t e n s i t y ( a . u . ) 2θ (degree) β (100) β (110) β (200) β (101) β (120) β (201) β (111) β (220) β (310) β (301) β (221) β (320) β (002) β (410) α (101) α (201) α (102) α (210) S D U T
3、定量分析：把α-Si,N看成q相，β-Si,N看成a相；则查PDF卡可得，Kβ=1.23，Kα=0.87，s为刚玉相，取(101)面（各自最强线）。那么：Kβ=Kβ/K,α=1.23/0.87=1.414;Ip/la=Kβ-(wp/wa),Wp+wa=l;Ip/la=4818/380;Wp=1/(1+K.β-Ia/Ip)=1/(1+1.414 X0.0789)=0.899;Wa=1-Wp=0.101所以：β-Si,N4含量： 89.9%α-Si,N4含量： 10.1%
3、定量分析： 把α-Si3N4看成q相，β-Si3N4看成a相；则 查PDF卡可得，Ks β=1.23，Ks α=0.87，s为刚玉相，取 (101)面（各自最强线）。那么 ： Kα β=Ks β /Ks α=1.23/0.87=1.414； Iβ /Iα =Kα β ·(wβ /wα )，wβ+wα =1； Iβ /Iα =4818/380; wβ=1/(1＋Kα β ·Iα /Iβ)=1/(1+1.414×0.0789)=0.899； wα =1-wβ=0.101 所以： β-Si3N4含量：89.9% α-Si3N4含量：10.1% S D U T
4、半高宽法计算晶粒大小：烧温半高宽（°）峰位 (°)t&mHKL度(℃)t1111.33630.2D (111)=6. 091nm400t2201.35650.341D (220)=6. 402nmt1110.82830.274D (111)=9. 830nm600t2200.79650.355D (220)=10. 906nm1110.40230.274t+mD (111)=20. 246nm8002200. 46150.261t+mD (220)=18. 824nmKa0.89 x0.15406x 57.3D=Da)= 6.091nm1.336 × cos(30.2 / 2)Bv/2 cos 0（弧度）=1/57.3（弧度）1° =T /180注意：B1/2用弧度单位
煅烧温 度（℃） 半高宽（º） 峰位（º） t&m HKL 400 1.336 30.2 t 111 1.356 50.341 t 220 600 0.828 30.274 t 111 0.796 50.355 t 220 800 0.402 30.274 t+m 111 0.461 50.261 t+m 220 4、半高宽法计算晶粒大小： 1/2 cos K D B (111) 0.89 0.15406 57.3 6.091 1.336 cos(30.2 / 2) D nm 注意：B1/2用弧度单位。 1°=π/180（弧度）=1/57.3（弧度） D(111)=6.091nm D(220)=6.402nm D(111)=9.830nm D(220)=10.906nm D(111)=20.246nm D(220)=18.824nm S D U T
有上述数据可知：（1）在400℃-800℃范围，随着温度的升高，四方相Zr02的晶粒大小增大，说明其结晶性在增强：（2）在800℃-1000℃范围，开始四方相Zr02特征峰几乎消失，单斜相Zr02特征峰显著增强，说明此温度范围，发生了四方相到单斜相的相变。m-monoclineT-tetragonalm(111)m(111)E1000℃mm1(1T1)800°℃DPT(311)T(200)C600℃B400℃A300℃506010203040708020/)
有上述数据可知： （1）在400℃-800℃范围，随着温度的升高，四方相ZrO2的 晶粒大小增大，说明其结晶性在增强； （2）在800℃-1000℃范围，开始四方相ZrO2特征峰几乎消 失，单斜相ZrO2特征峰显著增强，说明此温度范围，发生了 四方相到单斜相的相变。 S D U T