Expert: Paul Klarreich - 3/9/2007

Question
I do not understand this problem after nearly two days I have thrown in the towel.
The problem reads:
Radioactive Decay: Find the half-life of radioactive iodine (131I)if, after 20 days, 0.53 kilogram of an intial 3 kilograms remain.


Answer
Questioner:   Alicia
Category:  Advanced Math
 
Subject:  Logarithmics

>> Wasn't that the name of a hip-hop group?

Question:  I do not understand this problem after nearly two days I have thrown in the towel.

>> Dried it first, I hope.

The problem reads:
Radioactive Decay: Find the half-life of radioactive iodine (131I) if, after 20 days, 0.53 kilogram of an initial 3 kilograms remain.
..................................
Hi, Alicia,

Your basic equation here is:

M = M0 e^(-kt), where:

M is the mass present at time t.
M0 is the mass present at time  t = 0, which is normally an 'initial' time.
k is a constant that is not equal to, but related to, the half-life.

In your example, M0 = 3.  M = 0.53, and  t = 20.

0.53 = 3 e^(-k(20))

0.53 = 3 e^(-20k)

Now solve for k.  Take the ln of each side:

ln 0.53 = ln (3e^(- 20k))
ln 0.53 = ln 3 + ln e^(- 20k)
ln 0.53 = ln 3 - 20k
20k = ln 3 - ln 0.53
k = (ln 3 - ln 0.53)/20

Now fire up your calculator. Push a few buttons and:

k = 0.08667  and your equation is:

M = M0 e^(-0.08667 t)

Now we want a half-life?  That will be a value of t.  It will be THE

value of t when M = M0/2, the time after initialization when you have

half the initial amount.

M0/2 = M0 e^(-0.08667 t)

Solve for t:

1/2 = e^(-0.08667 t)

Take ln on both sides:

ln (1/2) = ln (e^(-0.08667 t))

- ln 2 = -0.08667 t

t = ln 2 / 0.08667

Push those buttons and you have:

t = 7.997126677384066494055998297505 or about 8 seconds.