Difference between revisions of "Spontaneous fission rate for sample of U-235 (1mg/cm^2)"
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<math>N = \frac{N_a}{A} m = \frac{6.022*10^{23} \frac{nuclei}{mol}}{235 \frac{g}{mol}} .91 * 10^{-3} g = 2.33*10^{18} nuclei</math> | <math>N = \frac{N_a}{A} m = \frac{6.022*10^{23} \frac{nuclei}{mol}}{235 \frac{g}{mol}} .91 * 10^{-3} g = 2.33*10^{18} nuclei</math> | ||
− | <math>R = \lambda N = 3.12*10^{-17} sec^{-1} \times 2.33*10^{18} nuclei = \textbf{72. | + | <math>R = \lambda N = 3.12*10^{-17} sec^{-1} \times 2.33*10^{18} nuclei = \textbf{72.70 nuclei}</math> |
Revision as of 16:00, 24 April 2008
Goal: Calculate the fission rate of the bomb grade
that has a mass of Procedure: Givens-
Procedure-