beta decay
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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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