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