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Cesium-137 is part of the nuclear waste produced by uranium-235 fission. The half-life of cesium-137 is 30.2 years. How much time is required for the activity of a sample of cesium-137 to fall to 9.32 percent of its original value

User Jack Casey
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Answer:

t(9.32% remaining) = 203 yrs (3 sig. figs.)

Step-by-step explanation:

All radioactive decay follows a 1st order decay profile. This is defined by the expression ...

A =A₀e^-k·t

A = final activity = 9.32%

A₀ = initial activity = 100%

e = base of natural logs

k = rate constant = 0.693/t(1/2) = (0.693/30.2) yrs⁻¹ = 0.023 yrs⁻¹

t = time of decay = ln(A/A₀)/-k = ln(9.32%/100%)/-0.023 yrs⁻¹

= 203.286637 yrs (calc. ans.)

≅ 203 yrs (3 sig. figs.)

User Jeremy Cowles
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