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*10^{8} year \times 3.1536*10^{7} sec = 2.2195*10^{16} sec</math>
+
   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 \times 10^{-3} g</math>
+
           <math> m = .91 \cdot 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 \cdot 10^{16} sec} = 3.12 \cdot 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 \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*10^{-17} sec^{-1} \times 2.33*10^{18} nuclei = \textbf{72.696  nuclei}</math>
+
                <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 [math]U^{235}[/math] that has a mass of [math].91*10^{-3}[/math] Procedure: 

 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 \cdot 10^{-3} g[/math]

 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 \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 = 72.70  fissions/sec[/math]

 Probability of fission per decay = [math]2.0 \cdot 10^{-7} % =\gt  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]
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