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

Revision as of 15:59, 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]

 Procedure- [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{72.696  nuclei}[/math]

@@ Line 1: / Line 1: @@
-Goal: Calculate the fission rate of a bomb grade U235 that has a mass of .91*10^-3 g.
+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>
-Given -
+          <math> m = .91 \times 10^{-3} g</math>
-T1/2 = 7.038 * 10^8 year =>7.038 * 10^8 year * 3153600 seconds/year = 2.2195 * 10^16 sec
-m = .91 * 10^-3 g
-Calculations -
-<math>N_o = \frac{N_a}{A} * m = \frac{6.022*10^{23} nuclei/mol}{235 g/mol}* (.91 * 10^{-3} g) = 2.33 * 10^{18} nuclei</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>
-The quadratic formula is <math>-b \pm \sqrt{b^2 - 4ac} \over 2a</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>\sum_{n=0}^\infty \frac{x^n}{n!}</math>
+             <math>R = \lambda N = 3.12*10^{-17} sec^{-1} \times 2.33*10^{18} nuclei = \textbf{72.696  nuclei}</math>