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A 25.0 mg sample of 148/58 CE (t 1/2=48s) undergoes beta decay. After 4.50 min, how much of the original sample is left, what is the product formed and how much of the product is formed?

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Solution Summary

Beta decay is investigated. The solution discusses what is the product formed and how much of the product is formed.

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If it loses an electron Ce-148 becomes Pr-148, or in the way you write it 148/59 Pr.

Use this equation to find the rate constant:

ln(2) = kt{1/2}

k = ln(2)/t{1/2} = 0.69321/48 s = ...

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