c10 JWBS043-Rogers September 13, 2010 11:26 Printer Name: Yet to Come
FIRST-ORDER KINETIC RATE LAWS 145
decay per second. You have observed arateof decay−dX/dtin the units (number
of atoms)/time, usually given as a frequency (s−^1 ). Decay causes a decrease in the
number of atomsX, which accounts for the minus sign in front of the rate. If you
take 200 billion atoms, you find twice as many radioactive events. That only stands
to reason. Radioactive decay is a random phenomenon. Observing twice as many
atoms, one expects to see twice as many events in a specified time interval. Takingk
times as many atoms givesktimes as many radioactive events: rate=kX. Now you
have arate,arate law, and arate equation. The rate of disintegration is
−
dX
dt
=kX
The rate of product production is equal but opposite in sign to the rate of decay.
Provided that the daughter is not radioactive, starting with a number of atomsX 0 at
a timet=t 0 =0 (an arbitrary time that you choose to start watching your counter),
the average count rate gradually decreases with time. If you watch long enough,^1 the
count rate will be reduced to half of what it was att 0. This is called thehalf-timeof
the radioactive element.
The rate equation can be rearranged to give
dX
X
=−kdt
The ratiodX/Xbeing unitless,−kdtmust be unitless as well, so the unit ofkis s−^1
becausetis in seconds. Integration betweent 0 andtis straightforward:
∫t
t 0
1
X
dX=−k
∫t
t 0
dt
gives
ln
X
X 0
=−k(t−t 0 )
In exponential form
X
X 0
=e−k(t−t^0 )
X 0 is the initial concentration (a constant) andt 0 is usually set to zero so that
X 0 =X(0) and
X=X 0 e−kt
(^1) About 1600 years in the case of radium.