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 \cdot 10^{23} \frac{nuclei}{mol}}{235 \frac{g}{mol}} .91 \cdot 10^{-3} g = 2.33 \cdot 10^{18} nuclei</math> | <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 \cdot 10^{-17} sec^{-1} \times 2.33 \cdot 10^{18} nuclei = \ | + | <math>R = \lambda N = 3.12 \cdot 10^{-17} sec^{-1} \times 2.33 \cdot 10^{18} nuclei = 72.70 fissions/sec</math> |
+ | |||
+ | Probability of fission per decay = <math>2.0 \cdot 10^{-7} % => 2.0 \cdot 10^{-9} \times 72.70 fissions/sec</math> | ||
+ | |||
+ | <math>= 1.5 \cdot 10^{-5}%</math> chance there will be a fission each second ∴ <math>\textbf{not probable}</math> |
Latest revision as of 15:37, 30 April 2008
Goal: Calculate the fission rate of the bomb grade
that has a mass of Procedure: Givens-
Calculations-
Probability of fission per decay =
chance there will be a fission each second ∴