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lead-210 decays to produce exclusively mercury-206 with a half-life of 22.3 yr. if the sample initially contains 5.15 g of lead-210, what mass (in g) of lead-210 will be present after 13.7 years?

User TBrenner
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1 Answer

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Final answer:

To find the remaining mass of lead-210 after 13.7 years, we use the formula for exponential decay with the initial mass of 5.15 g and a half-life of 22.3 years, resulting in approximately 3.94 grams of lead-210 remaining.

Step-by-step explanation:

The question is asking to determine the remaining mass of lead-210 after 13.7 years given that lead-210 has a half-life of 22.3 years and the initial mass of 5.15 grams. To calculate the remaining mass, we can use the formula for exponential decay, which describes how the quantity of a radioactive substance decreases over time. The formula is:

N = N_0 (1/2)^(t/T)

Where:

  • N is the final amount of the substance
  • N_0 is the initial amount of the substance
  • t is the time that has passed
  • T is the half-life of the substance

Using the given values, N_0 = 5.15 g, t = 13.7 years, and T = 22.3 years, we substitute these into the equation to calculate N:

N = 5.15 g (1/2)^(13.7 years / 22.3 years)

Calculating this gives us the remaining mass of lead-210:

N ≈ 5.15 g (1/2)^(0.6143)

N ≈ 5.15 g (0.765)

N ≈ 3.94 g

Therefore, after 13.7 years, approximately 3.94 grams of lead-210 will be present in the sample.

User EBarr
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