A certain radioactive isotope is a by-product of some nuclear reactors. Due to an explosion, a nuclear reactor experiences a massive leak of this radioactive isotope. Fortunately, the isotope has a very short half-life of days. Estimate the percentage of the original amount of the isotope released by the explosion that remains days after the explosion.

73.5%

Step-by-step explanation:

The question is incomplete, the complete question is shown in the image attached.

Half life (t1/2) = 9 days

Time taken (t) = 4 days

From;

0.693/t1/2 = 2.303/t log (No/N)

Substituting values;

0.693/9 = 2.303/4 log(No/N)

0.693/9 × 4/2.303= log(No/N)

2.772/20.727= log(No/N)

Antilog (2.772/20.727)=No/N

1.36 = No/N

Therefore;

N/No = (1.36)^-1

N/No = 0.735

Percentage remaining after 4 days= 0.735× 100= 73.5%

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