Question

Phosphorus-32 has a half-life of 14.0 days. Starting with 6.00g of 32P, how many grams will remain after 98.0days ?

Answer 1

How many half-lives have passed in 98 days?

Answer 2

49?

Answer 3

if the half-life is 14 days, in 98 days, \(\sf \dfrac{98~days}{14 ~days/half-life}=7~half-lives\) have passed.

Answer 4

If we know the substance decreases by half every half-life, then we can use \(A_t=A_o(\dfrac{1}{2})^n\) to determine the amount left (\(A_t\)) after \(n\) half-lives.

Or if you know a bit more math, it would be good to get acquainted with exponential decay. First-order integrated rate law:

\(\large ln[A]_t=-kt+ln[A]_o\)

where \([A]_o\) is the initial concentration of the substance
\(k\) is the decay constant
\(t\) is time
\([A]_t\) is the concentration of substance at time \(t\)

(Can also be written as \([A_{t}]=[A_0]*e^{-kt}\))

Answer 5

(6.00 g) x (1/2)^(98.0/14.0) = 0.0469 g left