Asked by mary
5 mg iodine 131 its half life is 8.1 days what is the time when 1% remain
Answers
Answered by
DrBob222
k = 0693/t<sub>1/2</sub>
Solve for k and substitute into the equation below.
ln(No/N) = kt
Use 5mg for No.
Use 1% x 5 = 0.05 for N.
k from above.
Solve for t.
t will be in days if you used 8.1 days for the half-life.
Solve for k and substitute into the equation below.
ln(No/N) = kt
Use 5mg for No.
Use 1% x 5 = 0.05 for N.
k from above.
Solve for t.
t will be in days if you used 8.1 days for the half-life.
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