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Showing posts with label Differential Equations. Show all posts
Showing posts with label Differential Equations. Show all posts

Wednesday, May 13, 2015

Exponential Decay Problems, 3

Category: Chemical Engineering Math, Differential Equations

"Published in Newark, California, USA"

How old is a bottle of wine if the tritium 3H content is 45% of a new wine? The half-life of tritium is 12.5 years.

Solution:

From the description of a given problem, it is about exponential decay problem. The rate of change of a substance is directly proportional to the negative of its substance present. The working equation can be expressed as follows



where

x = amount of tritium at time t
t = decaying time of tritium
k = proportional constant for decaying

By separation of variables, transfer x to the left side of the equation and dt to the right side of the equation as follows



Integrate on both sides of the equation, we have







Take the inverse natural logarithm on both sides of the equation, we have





If x = x0 and t = 0 at the start, then the value of C is 







Hence, the particular solution of the working equation is



If x = ½ x0 at t = 12.5 years, then the value of k is 



Take natural logarithm on both sides of the equation, we have 





Hence, the complete working equation of the above equation is 



If x = 0.45 x0 as stated in the problem, therefore, the age of a bottle of wine that contains tritium is 




Take the natural logarithm on both sides of the equation,w e have




Tuesday, May 12, 2015

Exponential Decay Problems, 2

Category: Chemical Engineering Math, Differential Equations

"Published in Vacaville, California, USA"

The half-life of Sr-90 is 29 years. What fraction of the atoms in a sample of Sr-90 would remain in 100 years later?

Solution:

From the description of a given problem, it is about exponential decay problem. The rate of change of a substance is directly proportional to the negative of its substance present. The working equation can be expressed as follows



where

x = amount of Sr-90 at time t
t = decaying time of Sr-90
k = proportional constant for decaying

By separation of variables, transfer x to the left side of the equation and dt to the right side of the equation as follows



Integrate on both sides of the equation, we have







Take the inverse natural logarithm on both sides of the equation, we have





If x = x0 and t = 0 at the start, then the value of C is






Hence, the particular solution of the working equation is
 
 

If x = ½ x0 at t = 29 years, then the value of k is




Take natural logarithm on both sides of the equation, we have





Hence, the complete working equation of the above equation is




If t = 100 years, therefore, the fraction of Sr-90 in a sample is