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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) If the average molecular mass of compounds in food is 50.0 g, how many molecules are mere in 1.00 kg at food? (b) How many ion pairs are created in 1.00 kg of food, if it is exposed to 1000 Sv and it takes 32.0 eV to create an ion pair? (c) Find the ratio of ion pairs to molecules. (d) If these ion pairs recombine into a distribution of 2000 new compounds, how many parts per billion is each?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.
- - 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 1970.059X= 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…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 - (-)]
- 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.Eg Q.1/ Prove that: ni = /NcNy Using Eg Ec- Ev %3D Solution: n.p= ni? I Page L ot 1 Words 15 SEngish uS2. Show that the coefficient of spontaneous emission Amt is given by Amt = 8n he (v mt) Bmf ? %3D
- 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