Question

Phosphorus-32 (P-32) has a half-life of 14.2 days. If 450 g of this substance are present initially, find the amount Q(t) present after t days. (Round your growth constant to four decimal places.) Q(t) =

What amount will be left after 17.6 days? (Round your answer to three decimal places.)

Answer 1

half life of phosphorus -32 = 14.2 days

initial amount = 450 g

so, standard exponential function is written as

A = Ao e^kt

plugging A o = 450 , A = 450/2 , t = 14.2

and finding the value of k

450/2 = 450 e^(14.2k )

dividing both sides by 450

1/2 = e^(14.2k )

taking natural log on both sides

ln (1/2) = 14.2k ln e

ln e = 1

so

ln (1/2 ) = 14.2 k

k = ln (1/2)/ 14.2

k = -0.0488

hence, Q(t) peresent after t days is

Q(t) = 450 e^(-0.0488 t )

amount after 17.6 days

plugging t = 17.6 in the equation of Q(t)

Q(t) = 450 e^(-0.0488* 17.6 )

Q(t) = 450 ( 0.4235)

= 190.592

amount left after 17.6 days = 190.592 g