The half-life of krypton-91 ^(91Kr) is 10 seconds. At time

Question

t = 0
a heavy canister contains 6 g of this radioactive gas.
(a) Find a function
m(t) = m02^−t/h
that models the amount of ^(91)Kr remaining in the canister after t seconds.
m(t) =

(b) Find a function
m(t) = m0e^−rt
that models the amount of ^(91)Kr remaining in the canister after t seconds.
m(t) =

(c) How much ^(91)Kr remains after one minute? (Round your answer to three decimal places.)
g

(d) After how long will the amount of ^(91)Kr remaining be reduced to 1 μg (1 microgram, or 10^−6 g)? (Round your answer to the nearest whole number.)
sec

Answer 1

since half-life is given, use 1/2 as a base

so m(t) = 6 (1/2)^(t/10), where t is number of seconds

b) using base e .....
m(t) = 6 e^(kt), where k is a constant
We have to find k
given, when t = 10, m(10) = 3
3 = 6 e^(10k)
.5 = e^(10k)
ln .5 = 10k
k = -.069315

m(t) = 6 e^(-.069315t)

c) if t = 1 minute = 60 seconds
using first equation:
m(60) = 6 (1/2)^6 = .09375

using 2nd equation:
m(60) = 6 e^(-.069315(60)) = .09375

d) 10^-6 = 6 (1/2)^(t/10)
.000001666... = (.5)^(t/10)
t/10 = 19.1646..
t = appr 191.95 seconds
or appr 3.2 minutes

I will leave it up to you to use the second equation to obtain the same answer.