I've a question from

a sample of radioactive iodine-133 gave with a geiger counter 3150 counts per minute at a certain time and 3055 counts per minute exactly 1 hour later calculate the half life period of iodine-133

me

22.06 Hours

We are given;

Radioactive isotope Iodine-133

Initial counts per minute = 3150 counts

Counts after decay = 3055 per minute

Time taken = 1 hour

But, we know that to calculate the remaining amount of the radioactive isotope, then,

Remaining amount = Initial amount × 0.5^n , where n is the number of half-lives.

Therefore;

3055 counts = 3150 × 0.5^n

0.5^n = 0.9698

To get n

n = log 1 ÷ log 0.9698

 = 0.044

But, n is given by dividing time and the number of half lifes

Therefore, half life = time ÷ Number of half lives

                              = 1 hour ÷ 0.044

                              = 22.60 hours

Therefore, the half life of iodine-133 is 22.60 hours

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