High School

Solve for [tex]n[/tex] in the formula for the half-life of a radioactive isotope:

[tex]A_{2} = \frac{A_{1}}{2^n}[/tex]

Given: [tex]A_{1} = 3700[/tex] and [tex]A_{2} = 200[/tex].

Answer :

To solve for the number of half-lives (n), substitute the initial and final activities into the half-life formula and take the logarithm. For the given activities of 3700 and 200, approximately 4.216 half-lives are needed for the activity to reduce from its initial value to 200.

To solve for the number of half-lives (n) required for a given amount of a radioactive isotope to reduce from its initial activity (A1) to its activity at some time later (A2). We have A1 = 3700 and A2 = 200. According to the provided half-life formula A2 = (A1)/(2n), we need to find n.

Substituting the given values into the formula:

  1. 200 = 3700/(2n)
  2. 2n = 3700/200
  3. 2n = 18.5
  4. Now we take the logarithm base 2 of both sides: log2(2n) = log2(18.5)
  5. n = log2(18.5)
  6. Using a calculator, we find that n \\approx 4.216

So approximately 4.216 half-lives are needed for the activity to decrease from 3700 to 200.

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