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

Goal: Calculate the fission rate of a bomb grade U235 that has a mass of .91*10^-3 g.

Procedure:

Given -

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]\displaystyle{ 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]

The quadratic formula is [math]\displaystyle{ -b \pm \sqrt{b^2 - 4ac} \over 2a }[/math]

[math]\displaystyle{ \sum_{n=0}^\infty \frac{x^n}{n!} }[/math]