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- The Csl crystal structure is BCC. The equilibrium spacing is approximately r0=0.46nm . If Cs+ ion a occupies a cubic volume of r03, what is the distance of this ion to its "nearest neighbor” I+ ion?(a) Show that if you assume the average nucleus is spherical with a radius r=r0A1/3, and with a mass at A u, then its density is independent at A. (b) Calculate that density in u/fm3 and kg/m3, and compare your results with those found in Example 31.1 for 56Fe.Suppose the range for 5.0 MeVa ray is known to be 2.0 mm in a certain material. Does this mean that every 5.0 MeVa a ray that strikes this material travels 2.0 mm, or does the range have an average value with some statistical fluctuations in the distances traveled? Explain.
- 4. From van der Waals equation, using (3²0) T (2²) ₁ = 0 Ꭲ a = ²-RTcVmc 8 Show that = and Tc = b = 8a 27Rb ; Vmc 3 5. Show that the Lennard-Jones energy (i) Pc = a 276² ---(ii) has a minimum value equal to - at r = 2¹/66. Vmc = 3b 12 U₁₁ (r) = 4 € [¹ - (-)]- reg YEARY PC PB YD TEMP PRP Source df MS Number of obs 29 F(6, 22) = 1777.52 Model 2025.82114 337.636856 Prob > F 0.0000 = Residual R-squared Adj R-squared 4.17886221 22 .189948282 0.9979 0.9974 Total 2030 28 72.5 Root MSE = .43583 YEAR Coefficient Std. err. t P>|t| [95% conf. interval] Y .0640617 .0523215 1.22 0.234 -.0444465 .1725699 PC -.0179212 .010265 -1.75 0.095 -.0392094 .003367 PB .0058838 .0050092 1.17 0.253 -.0045047 .0162722 YD .0978191 .0142261 6.88 0.000 .068316 .1273222 TEMP .0021932 .0054143 0.41 0.689 -.0090354 .0134218 PRP .0087202 .0091906 0.95 0.353 -.01034 .0277804 cons 1966.475 1.727965 1138.03 0.000 1962.892 1970.059Eg Q.1/ Prove that: ni = /NcNy Using Eg Ec- Ev %3D Solution: n.p= ni? I Page L ot 1 Words 15 SEngish uS
- 2. Show that the coefficient of spontaneous emission Amt is given by Amt = 8n he (v mt) Bmf ? %3DX= 5009618.596 Y=3664880.325 Z=1409190.807 SAT ID, X, Y, Z1 -16344666.151 -2728939.557 20658479.4432 14599389.291 21642189.782 -4841358.4083 -19782593.480 8031126.013 15713919.0265 10606629.037 11640082.198 -21490132.2286 5977034.172 25070732.986 6414596.3387 -10188818.151 13104378.593 -20387747.4068 -25478269.471 -6127327.576 -4840854.5699 -19660053.951 8957811.816 -15504101.90910 9176950.240 -23038750.733 9288715.55811 -21607341.598 -4784060.713 13988573.89712 19146698.015 -4291564.379 17616168.31313 20703829.603 12802029.889 -10783033.52914 -6803941.208 -15979931.468 20444839.08815 26544908.019 2713833.024 -2471391.08816 -9064955.452 -11862746.652 -22100079.35817 -3913538.697 14423991.316 22362387.48918 -17734815.436 -12799604.808 14692010.24119 6629923.708 15395724.715 20282174.52320 14081250.296 -22413255.230 -1110279.75221 6774854.395 -16745700.078 -18696887.44222 -17987713.323 391728.476 19756616.90823 -25791816.315 2079256.748 -6572981.34524 18096088.396 7150727.508…92: Use the gomsiom trial function You = Ae-bx aupper bound for the gromdstate the following potontin Ca) : Vcx) - dx Vax) =dXF Ans (a) 3 d2 s to estimate the lowest 3.
- 12) Fill in the missing values. Ffrict= μ = 0.1 8 = m = 5 kg Fnet Fnorm = Fgrav 11 ...v FfrWhat is the Sl equivalent of 4 Ci? a. 150,000,000,000 or 1.48 X 10(12) Bq Ob. 0.04 Bq c. 0.001032 C/kg Od. 148,000,000,000 or 1.48 X 10(11) Bq7. If the radius of the nucleus R = 1.2x 10^-13 cm, and known it is 1 fermi = 10^-13 cm, the nuclear radius of the 12C carbon atom is a. 1.2 fermi b. 2.7 fermi c. 3.6 fermi d. 1.7 fermi e. 2.2 fermi