Calculating Spontaneous Fission Rate of 238U and Its Half-Life

AI Thread Summary
The discussion centers on calculating the spontaneous fission rate of 238U, which is stated as one fission per 100 seconds per gram, and its relation to the half-life of 5.5 x 10^15 years. A participant attempts to use Avogadro's number and the mass of 238U to derive the fission rate but encounters difficulties. The key formula referenced is N = N_0 * 2^(-t/T), where N_0 represents the initial number of atoms in one gram. The conversation highlights the challenges in applying this formula correctly to demonstrate the relationship between fission rate and half-life. Overall, the thread emphasizes the complexity of calculating nuclear decay rates and the need for precise application of formulas.
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Homework Statement



Spontaneous fission rate of 238U is one fission per 100s per g show that it is equal to half life of 5.5x10^15year

Homework Equations





The Attempt at a Solution



I tried so many different ways, one of them was to use avogadro number with weight mass 238 and then multiply it by the fission rate but it does not work, any idea
 
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If N_0 is number of atoms in 1 g.

The main formula is

<br /> N=N_02^{-t/T}<br />

After t=100 s we get

<br /> N_0-1=N_02^{-t/T}<br />
 

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