Difference between revisions of "Spontaneous fission rate for sample of U-235 (1mg/cm^2)"
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Calculations - | 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> | |
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+ | The quadratic formula is <math>-b \pm \sqrt{b^2 - 4ac} \over 2a</math> | ||
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+ | <math>\sum_{n=0}^\infty \frac{x^n}{n!}</math> |
Revision as of 15:08, 24 April 2008
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 -
The quadratic formula is