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Data Provided: A formula sheet and table of physical constants is attached to this paper. DEPARTMENT OF PHYSICS AND ASTRONOMY Spring Semester (2013-2014) NUCLEAR PHYSICS 2 HOURS Answer question ONE (Compulsory) and TWO other questions. All questions are marked out of ten. The breakdown on the right-hand side of the paper is meant as a guide to the marks that can be obtained from each part. Please clearly indicate the question numbers on which you would like to be examined on the front cover of your answer book. Cross through any work that you do not wish to be examined. (2) (d) The Po isotope is cases yields: 19.54 MeV. 210 84 Po , Substituting in using the values above for the two (e) Number of nuclei = 1 x 10-6 x Avagadro / 210 = 2.87 x 1015. The decay rate lamda = ln2/half life = 0.693/ 138 x 24 x 3600 = 5.81 x 10-8 s-1 Power 1.38 x 10-4 Watts (3) (d) Substituting in yields 0.36 for U-234 and 0.96 for C-12, see spreadsheet calculation: (4) (c) r = φ σ n V (d) σ = r / ( φ n V ) total number of interactions of 0.032 s−1. The number of hydrogen atoms is 2 × 400 × 1000 × NA/18 Therefore, the cross-section = 9.2×10−48 m2. (e) Use path length = 1/(n x cross sec) where n is the proton density in the Sun = 1.4 x 103 kg/m3 / 1.67 x 10-27 kg. = 8.4 x 1029. ANS: 1.3 x 1017 m [5] (d) To get the nuclear radius use 1.2 fm x A1/3 Hence R2 = (1.2 fm)2 x A2/3. For N-15 the j = 1/2 and for O-17 j = 5/2 Substituting gives: Q for 17O = -0.0326, for 15 N=0