determine the amount present after 54 minutes and after 2 hours. when will the amount present be less than 2 kg

Question

The radioactive isotope U-24 has a half-life of 18 minutes. If there's presently 150kg of U-24: determine the amount present after 54 minutes and after 2 hours. when will the amount present be less than 2 kg

Answer 1

An exponential decay, where we know the half-life (the time it takes for the amount to be reduced by half) can be respresented with an equation:

 

    (Final Amount) = (Initial Amount)*(1/2)^(time/half-life)

 

    A_f = Ai*(1/2)^(t/t_1/2) 

 

The half-life is 18 minutes, so:

 

   A_f = A_i*(1/2)^(t/18)     where t is measured in minutes

 

After 54 minutes:

 

    A_f = (150kg)*(1/2)^(54/18)

 

    A_f = 18.75 kg

 

After 2 hours (120 minutes):

 

    A_f = (150kg)*(1/2)^(120/18)

 

    A_f = 1.48 kg (rounded to three significant figures)

 

To figure out how much long it takes for there to be 2kg remaining, we plug in 2kg for "A_f", and use logarithms to solve for "t"

 

    2 = (150)*(1/2)^(t/18)

 

    2/150 = (1/2)^(t/18)

 

Let's take the natural log (or any other base log) of both sides:

 

    ln(2/150) = (t/18)*ln(1/2)

 

    ln(2) - ln(150) = (t/18)*ln(1/2)

 

And we will now solve for "t":

 

    (ln(2) - ln(150))/ln(1/2) = t/18

 

    t = 18*(ln(2) - ln(150))/ln(1/2)

 

    t = 112 minutes, again rounded to three digits

 

I hope this helps!