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Unformatted text preview: MP200  Radiation Physics Fall 2010 Assignment 02 Due 09/22/10 ( Show your work for each problem. Partial credits are awarded.) 1. ( Problem # 02 on page 18 of Attix ) The following set of counts readings was made in a gradientfree ray field, using a suitable detector for repetitive time periods of one minute: 18500; 18410; 18250; 18760; 18600; 18220; 18540; 18270; 18670; 18540. (a) What is the mean value of the number of counts? (b) What is its standard deviation (S.D)? (c) What is the theoretical minimum S.D. of the mean? (d) What is the actual S.D. of a single reading? (e) What is the theoretical minimum S.D. of a single reading? Solution: (a) mean = N i 10 = 184760 10 = 18476 (01 point) (b) standard deviation of the mean = = n = 1 n 1 10 X 1 ( N i N ) 2 1 2 = v u u t 1 9 305840 = 184 . 2 = 184 . 2 10 = 58 . 2 1 (03 points) (c) min = min n min = N = 18476 = 135 . 92 min = 135 . 92 10 = 42 . 98 (03 points) (d) Actual S.D. = 1 n 1 10 X 1 ( N i N ) 2 1 2 = 184 . 34 (02 points) (e) min = N = 18476 = 135 . 92 (01 point) 2. A point source of 60 Co gamma rays emits equal numbers of photons of 1 . 17 and 1 . 33 MeV , giving a flux density of 3 . 7 10 9 photons.cm 2 .sec 1 at a specified location. What is the energy flux density there, ex pressed in erg.cm 2 .s 1 and in J.m 2 .min 1 ? Flux density due to each energy, = 5 . 7 2 10 9 photons.cm 2 .s 1 = 2 . 85 10 9 p.cm 2 .s 1 = 2 . 85 10 13 photons.m 2 .s 1 2 Energy density is given by, = E 1 . 17 MeV 1 = 1 . 17 10 6 ( eV/photons ) 2 . 85 10 13 ( photons.m 2 .s 1 60( s.hr 1 ) 1 . 602 10 19 ( J.eV 1 ) (04 points) 1 . 33 MeV 2 = 1 . 33 10 6 ( eV/photons ) 2 . 85 10 13 ( photons.m 2 .s 1 60( s.hr 1 ) 1 . 602 10 19 ( J.eV 1 ) (04 points) Total = 1 + 2 (01 point) = 685 J.m 2 .min 1 (b) 1 J = 10 7 erg = 685 10 7 erg 10 4 ( cm 2 ) 60 s = 11416 . 6 erg.cm 2...
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 Fall '10
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