Nitrogen-13 is used in tracers injected into the bloodstream for positron emission tomography (PET). The half-life of nitrogen-13 is 10.0 minutes. How much time is required for the activity of...

Nitrogen-13 is used in tracers injected into the bloodstream for positron emission tomography (PET). The half-life of nitrogen-13 is 10.0 minutes. How much time is required for the activity of a sample of nitrogen-13 to fall to 17.8 percent of its original value?

Expert Answers
gsenviro eNotes educator| Certified Educator

Half-life is the time period in which a substance reduces to 50% of its original value. Thus, a substance reduces to 50% original value in 1 half-life, to 25% in 2 half-lives, 12.5% in 3 half-lives and so on. Here, nitrogen-13 has a half-life of 10 minutes. That means, 50% of nitrogen-13 decays in 10 minutes. 

The number of half-lives can be calculated by using the following expression:

2^n = 100 / (% of original content left)

where, n is the number of half-lives. Here, 17.8% of the original content is left. 

Therefore, 2^n = 100/17.8 

or n log 2 = log (100/17.8)

solving this equation, we get, n = 2.49

That is, in 2.49 half-lives, a substance will decay to 17.8% of its original content. 

Since the half-life of nitrogen-13 is 10 minutes, it will take 24.9 min (= 2 half lives x 10 min per half life) for it to decay to 17.8% of its original value.

Hope this helps. 

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