HalfLife - -1.58 300 2.40E+03 1.89E-01-1.67 330 2.20E+03...

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0 100 200 300 400 500 600 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 f(x) = -0.01x - 0.12 Linear Fit Ln(Normalized Count Rate) vs. Time(sec) Column E Linear Regression for Column E Ln(Normalized Count Rate) Time (sec)
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50 150 250 350 450 550 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 f(x) = 0.88·0.99^x Nonlinear Fit NCR vs. Time (sec) Column D Exponential Regression for Column D Normalized Count Rate Time (sec)
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Finding the Half Life of A Given: Decay of Al27 Al27→Al28→Si28 Time(sec) Count Rate(cps) Normalized Count Rate Natural Log of NCR ΔS=30(sec) 0 1.27E+04 1 0 30 8.26E+03 6.50E-01 -0.43 60 8.26E+03 6.50E-01 -0.43 90 6.75E+03 5.31E-01 -0.63 120 6.73E+03 5.30E-01 -0.64 150 5.50E+03 4.33E-01 -0.84 180 4.19E+03 3.30E-01 -1.11 210 2.95E+03 2.32E-01 -1.46 240 3.54E+03 2.79E-01 -1.28 270 2.62E+03 2.06E-01
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Unformatted text preview: -1.58 300 2.40E+03 1.89E-01-1.67 330 2.20E+03 1.73E-01-1.75 360 1.97E+03 1.55E-01-1.86 390 1.58E+03 1.24E-01-2.08 420 1.08E+03 8.50E-02-2.46 450 1.02E+03 8.03E-02-2.52 480 1.07E+03 8.43E-02-2.47 510 9.08E+02 7.15E-02-2.64 540 6.43E+02 5.06E-02-2.98 570 6.61E+02 5.20E-02-2.96 600 4.03E+02 3.17E-02-3.45 N(t)=N e-t Al28 From the graph of Ln(Normalized Count Rate) vs. Time the decay constant is .0052(1/sec) From this equation, the half life of Al28 can be calculated From both graphes were exaclty the same constant proving that both a linear fit and nonlinear fit would be accurate T 1/2 =ln2/ T 1/2 =2.2216 mins...
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HalfLife - -1.58 300 2.40E+03 1.89E-01-1.67 330 2.20E+03...

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