Algebra ASAP

posted by Fay

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

  1. Reiny

    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.

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