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Iodine-131, a beta emitter, is one such product of nuclear fission, extremely dangerous because it is typically produced in huge quantities and because it is incorporated in the human body in the thyroid gland. The half-life of 1311 is 8.04 days. What fraction (expressed as a decimal number) of the 1311 produced in a nuclear accident would remain after 60.0 days?

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Answer:

0.00568

Step-by-step explanation:

Recall that the fraction of radioactive material left after a time (t) is obtained from the formula;

N/No = (1/2)^t/t1/2

Where;

N= mass of radioactive material left after a time t

No= mass of radioactive material initially present

t= time taken for N mass of radioactive substance to remain = 60.0 days

t1/2= half-life of the radioactive substance= 8.04 days

Note that the required fraction is N/No, hence;

N/No= (1/2)^60.0/8.04

N/No= (1/2)^7.46

N/No = 0.00568

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