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I've a question from chemistry...

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

Please help me out!!​

User Jitin
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1 Answer

4 votes

Answer:

22.06 Hours

Step-by-step explanation:

  • Half life is the time it takes for a radioactive material or isotope to decay by half of its initial amount.

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

User Seph Cordovano
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