CHM115_SCB_Lecture5_Problems

# CHM115_SCB_Lecture5_Problems - cone pine that is 3000 y old...

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Nuclear Kinetics dN/dt = counts per s (or min, hr, . ..) = slope of N vs t Rate equation dN/dt = -kN d[A]/dt = -k[A] Integrated rate equation ln{N o /N} = kt ln{[A o ]/[A]} = kt

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The Nuclear Half-life Rates are commonly given in half-lives, the time required for one-half of a nuclide to decay. ln (N o /N) = kt k = ln 2 / t ½ or t ½ = ln 2 / k ln(N o / ½N o ) = kt ½ ln(N o / ½N o ) = ln 2 = kt ½ ln 2 = kt ½
Example How long will a cobalt–60 source be useful if it can be used until the gamma rays it produces reach 70 percent of the original intensity? 60 27 Co decays with a half–life of 5.3 years to produce 60 28 Ni. k = (ln 2) / t 1/2 = 0.693 / t 1/2 = 0.693/5.3 y = 0.131 y -1 ln(N o /0.70N o ) = kt = 0.131 t t = (ln 1.43)/0.131 = 0.358 / 0.131 = 2.7 y 1. Calculate k ( reminder of rounding ) 2. Use the integrated form of the rate equation to find t

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Question #3 What percent of the original 14 C remains in the heartwood of a bristle

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Unformatted text preview: cone pine that is 3000 y old? 14 C → 14 N + e-t 1/2 = 5730 y ln{N o /N} = kt ln x = y y = e x ln 2 = kt 1/2 What percent of the original 14 C remains in the heartwood of a bristle cone pine that is 3000 y old? 14 C → 14 N + e-t 1/2 = 5730 y ln{N o /N} = kt ln x = y y = e x 69.6 % ln 2 = kt 1/2 Question #3 N.B.E. usually in energy per atom in eV or MeV 14 7 N → 7 1 1 H + 7 1 n 14.0031 amu 14.11543 amu N.B.E. = 1.68 x 10-11 J/nuclide 1 MeV/nuclide = 1.60 x 10-13 J/nuclide N.B.E. = 105 MeV Nuclear Binding Energy BE / nucleon = BE / nucleon = BE nucleon = Nuclear binding energy Mass number 1.009 x 10 10 kJ/mole N-14 14 mole p & n = 7.221 x 10 8 kJ/mole of nucleons 105 MeV/N-14 nucleus 14 nucleons/N-14 nucleus = 7.50 MeV / nucleon NBE: Clue to the type of nuclear reactions...
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CHM115_SCB_Lecture5_Problems - cone pine that is 3000 y old...

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