Radiation!!?

A paricular radio-isotope has a half life of 20 min. If the activity of a sample is 800 Bq at 1pm what will the time be when its activity has decayed to 100 BqRadiation!!?
I'm not going to solve the problem for you but I'm going to give you the process then plug in the figures.

Let :

dQ/dt = the rate of change in radioactivity which is proportional to radioactivity at any time. then;



dQ/dt = -kQ where k is the constant of proportionality. The rate of change in radioactivity is negative because is is decreasing or attenuating. Thus;

dQ/Q = -kdt

lnQ = -kt + lnC where lnC is the constant of integration;

lnQ - lnC = -kt

ln(Q/C) = -kt



when t = 0, Q = Qo where Qo is the initial rate of radioactivity

Thus;

Qo =C=800

ln(Q/800) = -kt

when t=20 minutes, Q = 400 Hence;

ln(400/800) = -k(20)

ln(1/2) = -20k

k = -ln(1/2)/20 thus;

ln(Q/800) = ln(1/2)t/20 therefore;

t = 20 ln(Q/800)/ln(1/2)

Now just substitute 100 for Q and you get the value of t, which is the number of minutes after 1PM when the radioactivity has decayed to 100Bq. That would be at 2PM. OPS! Why did I gave you the answer? Am I spoon feeding you?Radiation!!?
Hi,

A(t) = A(0) x 0.5^(t/HL)

Where:

A(t) = activity at time t = 100 Bq

A(0) = activity at zero time = initial activity = 800 Bq

t = time interval = ?

HL = Half Life = 20 min.

=%26gt; 100 = 800 x 0.5^(t/20)

100/800 = 0.5^(t/20)

0.125 = 0.5^(t/20)

taking log on both sides =%26gt;

log(0.125) = t/20 x log(0.5)

log(0.125)/log(0.5) = t/20

3 = t/20

or t = 3 x 20 = 60 min



This is your required answer. It can also be checked as

activity from 800 Bq to 400 Bq becomes in 20 mins, then from 400 Bq to 200 Bq becomes in another 20 mins (total 40 mins) and then from 200 Bq to 100 Bq in more 20 minutes (total 60 mins).



Hope to answer you well. Bye.Radiation!!?
well after 20mins the sample should be at 400Bq then another 20mins it'll be at 200Bq then another 20 mins it'll be at a hundred



that totals 1hour! so the time must be 2pm