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beta decay

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?

Solution Preview

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 = 0.0144 ...

Solution Summary

Beta decay is investigated.