Difference between revisions of "Spontaneous fission rate for sample of U-235 (1mg/cm^2)"
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Goal: Calculate the fission rate of the bomb grade <math>U^{235}</math> that has a mass of <math>.91*10^{-3}</math> | Goal: Calculate the fission rate of the bomb grade <math>U^{235}</math> that has a mass of <math>.91*10^{-3}</math> | ||
Procedure: | Procedure: | ||
− | Givens- <math> T_{1/2} = 7.038 | + | Givens- <math> T_{1/2} = 7.038 \cdot 10^{8} year \times 3.1536 \cdot 10^{7} sec = 2.2195 \cdot10^{16} sec</math> |
− | <math> m = .91 \ | + | <math> m = .91 \cdot 10^{-3} g</math> |
− | Calculations- <math>\lambda = \frac{\ln{2}}{T_{1/2}} = \frac{\ln{2}}{2.2195 | + | Calculations- <math>\lambda = \frac{\ln{2}}{T_{1/2}} = \frac{\ln{2}}{2.2195 \cdot 10^{16} sec} = 3.12 \cdot 10^{-17} sec^{-1}</math> |
− | <math>N = \frac{N_a}{A} m = \frac{6.022 | + | <math>N = \frac{N_a}{A} m = \frac{6.022 \cdot 10^{23} \frac{nuclei}{mol}}{235 \frac{g}{mol}} .91 \cdot 10^{-3} g = 2.33 \cdot 10^{18} nuclei</math> |
− | <math>R = \lambda N = 3.12 | + | <math>R = \lambda N = 3.12 \cdot 10^{-17} sec^{-1} \times 2.33 \cdot 10^{18} nuclei = \textbf{72.70 fissions/sec}</math> |
Revision as of 19:54, 24 April 2008
Goal: Calculate the fission rate of the bomb grade
that has a mass of Procedure: Givens-
Calculations-