Difference between revisions of "Spontaneous fission rate for sample of U-235 (1mg/cm^2)"

Revision as of 16:17, 24 April 2008

Goal: Calculate the fission rate of the bomb grade [math]U^{235}[/math] that has a mass of [math].91*10^{-3}[/math] Procedure:

 Givens- [math] T_{1/2} = 7.038*10^{8} year \times 3.1536*10^{7} sec = 2.2195*10^{16} sec[/math]

         [math] m = .91 \times 10^{-3} g[/math]

 Calculations- [math]\lambda = \frac{\ln{2}}{T_{1/2}} = \frac{\ln{2}}{2.2195*10^{16} sec} = 3.12*10^{-17} sec^{-1}[/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{72.70  nuclei}[/math]

@@ Line 5: / Line 5: @@
            <math> m = .91 \times 10^{-3} g</math>
-   Procedure- <math>\lambda = \frac{\ln{2}}{T_{1/2}} = \frac{\ln{2}}{2.2195*10^{16} sec} = 3.12*10^{-17} sec^{-1}</math>
+   Calculations- <math>\lambda = \frac{\ln{2}}{T_{1/2}} = \frac{\ln{2}}{2.2195*10^{16} sec} = 3.12*10^{-17} sec^{-1}</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>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{72.70  nuclei}</math>
+                <math>R = \lambda N = 3.12*10^{-17} sec^{-1} \times 2.33*10^{18} nuclei = \textbf{72.70  nuclei}</math>