Phosphorus-32 has a half-life of 14.0 days. starting with 4.00 g of 32p, how many grams will remain after 84.0 days ?

The final amount will be 0.0625 grams.

Solution: Radioactive decay obeys first order kinetics and the first order integrated rate law equation is:

where, is the initial amount and A is the final amount, k is the decay constant and t is the time.

First of all we calculate the decay constant from given half life using the equation:

Given half life is 14.0 days,

So,

Initial amount is given as 4.00 g and the time is 84.0 days. Let's plug in the values in the first order integrated rate law equation and calculate the final amount.

taking anti ln to both sides:

A = 0.0625

So, the remaining amount after 84.0 days will be 0.0625 grams.

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