View source for Spontaneous fission rate for sample of U-235 (1mg/cm^2)

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} % => 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>