Tag: radioactive decay

Linearizing an Exponential Function: Radioactive Decay

Using this data for the decay of a radioisotope, find its half life.

t (s)A (g)
0100
10056.65706876
20032.10023441
30018.18705188
40010.30425049
5005.838086287
6003.307688562

We can start with the equation for decay based on the half life:

   A = A_0 (\frac{1}{2})^\frac{t}{\lambda}  

 
where:
 A = \text{Amount of radioisotope (usually a mass)}  A_0 = \text{Initial amount of radioisotope (usually a mass)}   t = \text{time}  \lambda = \text{half life}

and linearize (make it so it can be plotted as a straight line) by using logarithms. 

Take the log of each side (use base 2 because of the half life):

  \log_2{(A)} = \log_2{  \left( A_0 (\frac{1}{2})^\frac{t}{\lambda} \right)} 

Use the rules of logarithms to simplify:

 \log_2{(A)} = \log_2{ ( A_0 )} + \log_2{  \left( (\frac{1}{2})^\frac{t}{\lambda} \right)}   

  \log_2{(A)} = \log_2{ ( A_0 )} +  \frac{t}{\lambda}  \log_2{   (\frac{1}{2}) }      

 \log_2{(A)} = \log_2{ ( A_0 )} +  \frac{t}{\lambda}  (-1)   

  \log_2{(A)} = \log_2{ ( A_0 )} -  \frac{t}{\lambda}       

Finally rearrange a little:

  \log_2{(A)} =  -  \frac{t}{\lambda}  +  \log_2{ ( A_0 )}      

  \log_2{(A)} =  -  \frac{1}{\lambda} t +  \log_2{ ( A_0 )}

Now, since the two variables in the last equation are A and t we can see the analogy between this equation and the equation of a straight line:

 \log_2{(A)} =  -  \frac{1}{\lambda} t +  \log_2{ ( A_0 )}        

and,

   y =  m x +  b       

where:

   y = \log_2{(A)}        

   m =  -  \frac{1}{\lambda}        

   x = t       

   b =  \log_2{ ( A_0 )}

So if we draw a graph with log₂(A) on the y-axis, and time (t) on the x axis, the slope of the line should be:

   m =  -  \frac{1}{\lambda}

Which we can use to find the half life (λ).

Radioactive Half Lives

Since we most commonly talk about radioactive decay in terms of half lives, we can write the equation for the amount of a radioisotope (A) as a function of time (t) as:

  A = A_0 (\frac{1}{2})^\frac{t}{\lambda} 

where:
 A = \text{Amount of radioisotope (usually a mass)}  A_0 = \text{Initial amount of radioisotope (usually a mass)}   t = \text{time}  \lambda = \text{half life}

To reverse this equation, to find the age of a sample (time) we would have to solve for t:

Take the log of each side (use base 2 because of the half life):

  \log_2{(A)} = \log_2{  \left( A_0 (\frac{1}{2})^\frac{t}{\lambda} \right)} 

Use the rules of logarithms to simplify:

 \log_2{(A)} = \log_2{ ( A_0 )} + \log_2{  \left( (\frac{1}{2})^\frac{t}{\lambda} \right)}   

  \log_2{(A)} = \log_2{ ( A_0 )} +  \frac{t}{\lambda}  \log_2{   (\frac{1}{2}) }      

 \log_2{(A)} = \log_2{ ( A_0 )} +  \frac{t}{\lambda}  (-1)   

 \log_2{(A)} = \log_2{ ( A_0 )} -  \frac{t}{\lambda}      

Now rearrange and solve for t:

 \log_2{(A)} - \log_2{ ( A_0 )} = -  \frac{t}{\lambda}      

 -\lambda \left( \log_2{(A)} - \log_2{ ( A_0 )} \right) = t      

  -\lambda \cdot \log_2{ \left( \frac{A}{A_0} \right)}  = t

So we end up with the equation for time (t):

  t = -\lambda \cdot \log_2{ \left( \frac{A}{A_0} \right)}

Now, because this last equation is a linear equation, if we’re careful, we can use it to determine the half life of a radioisotope. As an assignment, find the half life for the decay of the radioisotope given below.

t (s)A (g)
0100
10056.65706876
20032.10023441
30018.18705188
40010.30425049
5005.838086287
6003.307688562