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Iron-59 is used to study iron metabolism in the spleen. Its half-life is 44 days. How many days would it take a 28.0 g sample of iron-59 to decay to 3.50 g?

User Edhedges
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Final answer:

To calculate the decay time for Iron-59 from 28.0 g to 3.50 g, we determine that it takes 3 half-lives, or 132 days, since the half-life of Iron-59 is 44 days.

Step-by-step explanation:

The decay of a radioactive isotope is an exponential process that can be described by the half-life, which is the time required for half of the isotope to decay. For Iron-59, with a half-life of 44 days, we use the concept of half-lives to determine the time it takes for a sample to decay from 28.0 g to 3.50 g.

Calculation Steps:

  1. Determine the number of half-lives: How many times do we need to halve the initial mass to reach the final mass?
  2. Calculate the total time: Multiply the number of half-lives by the half-life duration of the isotope.

Using the formula N(t) = N0(1/2)^(t/T), where N(t) is the final amount, N0 is the initial amount, t is the time elapsed, and T is the half-life, we can solve for t given N0 = 28.0 g and N(t) = 3.50 g.

First, we find the number of half-lives:

1 half-life: 28.0 g → 14.0 g

2 half-lives: 14.0 g → 7.0 g

3 half-lives: 7.0 g → 3.5 g

Thus, it takes 3 half-lives to go from 28.0 g to 3.5 g.

Next, we multiply the number of half-lives by the half-life period of Iron-59 to find the total time:

3 half-lives × 44 days/half-life = 132 days

Therefore, it would take 132 days for a 28.0 g sample of Iron-59 to decay to 3.50 g.

User Pedro Oliveira
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