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 a bomb grade U235 that has a mass of .91*10^-3 g.
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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>
 +
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>
  
Procedure:
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          <math> m = .91 \cdot 10^{-3} g</math>
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  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>
  
Given -  
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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>
  
T1/2 = 7.038 * 10^8 year =>7.038 * 10^8 year * 3153600 seconds/year = 2.2195 * 10^16 sec
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                <math>R = \lambda N = 3.12 \cdot 10^{-17} sec^{-1} \times 2.33 \cdot 10^{18} nuclei = 72.70  fissions/sec</math>
         
 
m = .91 * 10^-3 g
 
  
Calculations -
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  Probability of fission per decay = <math>2.0 \cdot 10^{-7} % => 2.0 \cdot 10^{-9} \times 72.70 fissions/sec</math>
 
No = (Na/A) * m = ([6.022 * 10^23 nuclei/mol] / [235 g/mol])* (.91 * 10^-3 g) = 2.33 * 10^18 nuclei
 
  
λ = ln(2)/T1/2 =ln(2)/(2.2195 *10^16 sec) = 3.12 * 10^-17 sec-1
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                <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]