110666 points waneF MAC 7 2 3 053 The halflife of cobalt 60

11-0-666 points waneF MAC 7 2 3 053 The half-life of cobalt 60 is 5 years. (Round the decay constant to three significant digits.) (a) Obtain an exponential decay model for cobalt 60 in the form Q() Que Q(t) = (b) Use your model to predict, to the nearest year, the time it takes one fifth of a sample of cobalt 60 to decay yr

Solution

a)Q(t) = Q0e^kt.  
given cobalt half life 60 is 5 years
At half-life, Q(t)/Q0 = 0.5 =1/2
Put 0.5 = e^(kt.)

Take natural logs of both sides:

ln(0.5) = -kt * ln(e)
kt = 0.693 (since ln(0.5) = -0.693 and ln(e) = 1)

5 *k = 0.693 (since t=5years)
k = 0.693 /5 = 0.1386

So the model is: Q(t) = Q0e^kt.
Q(t) = Q0e^( 0.1386t)

b)At one-fifth decayed,

1/5 decays means that 4/5 remains
Q(t)/Q0 = 4/5=0.8
0.8 = e^( 0.1386t)
ln(0.8) = 0.1386t
-0.2231 = 0.1386t
t = 0.2231 / 0.1386
t = 1.6096
Say two years, to the nearest year.