NN_{o} et4.78 x 10^{16}e(0.263/yr) (12 yr)2.04 x 10^{15}AA_{o} etAA_{o}etlnAA_{o}ttlnAA_{o}tln0.001 Ci0.0108 Ci0.263 year^{1}t9.05 yearsAtomic and Nuclear PhysicsDOE-HDBK-1019/1-93RADIOACTIVITYRev. 0Page 35NP-01(c)The number of californium atoms that will remain in 12 years can be calculatedfrom Equation (1-4). (d)The time that it will take for the activity to reach 0.001 Ci can be determinedfrom Equation (1-5). First, solve Equation (1-5) for time.Inserting the appropriate values in the right side of this equation will result in therequired time.PlottingRadioactiveDecayIt is useful to plot the activity of a nuclide as it changes over time. Plots of this type can beused to determine when the activity will fall below a certain level. This plot is usually doneshowing activity on either a linear or a logarithmic scale. The decay of the activity of a singlenuclide on a logarithmic scale will plot as a straight line because the decay is exponential.

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