Question

A radioactive substance decays according to the formula A=A0e^kt

where

A0 is the initial amount of substance (in grams)
A is the amount of substance(in grams) after t years
k is a constant

The half-life of the substance is 10 years.If we begin with 20g of the substance,how much will be left after 5 years?

Answer 1

you can go the e^kt route, but it's more transparent to use base 2, since we're talking about half-lives. 2^-1 = 1/2, so

A = Ao*2^(-t/10)
after 5 years, you have

A = 20*2^(-5/10) = 14.14g

Of course, since 2 = e^ln2, that means you could use

(e^ln2)^(-t/10) = e^(-t*ln2/10) = e^(-0.0693t)

Answer 2

how much of the substance would be present after 5 years

Answer 3

e=1/2 since we are talking about half-lives
AO=20g
constant,k=elapsed time(t)/half-life
=5 years/10 years
=1/2
IF A=AOe^1/2
=20g(1/2)^1/2
=14.14 g
thats the solution happy dance

Answer 4

Thank you so much you helped