a) For a sample of Strontium-90 with a half-life of 28.1 years, how many years will it take for the sample to decay to 12.5% of its original mass?

b) A sample of Ge-66 with a half-life of 2.5 hours was left for 10 hours, and only 5 grams remain. What was the original mass of the sample?

c) If a 100-kg sample of Element X decays to 12.5 kg in 1 day, what is the half-life of Element X in hours?

d) Strontium-90 is used as a radioactive tracer with a half-life of 28.1 years. How many atoms out of a [tex]$1.68 \times 10^{23}$[/tex] atom sample will remain undecayed after 56.2 years?

e) What percentage of a 900-gram sample of Potassium-37, with a half-life of 1.23 seconds, exists after 6.15 seconds?

This detailed answer provides calculations and formulas to solve for different scenarios involving half-life. It gives step-by-step explanations and examples for each question.

Explanation:

a)

For a sample of Strontium-90 with a half-life of 28.1 years, it will take 56.2 years for the sample to decay to 12.5% of its original mass. This can be calculated by dividing 28.1 years (the half-life) by 0.5 (the fraction remaining) and then multiplying by 2 (since the decay process occurs twice within 56.2 years).

b)

For a sample of Ge-66 with a half-life of 2.5 hours, if only 5 grams remain after 10 hours, the original mass of the sample can be calculated using the formula:

Original Mass = (Final Mass) × 2^(Number of Half-Lives)

Plugging in the values, we have Original Mass = 5 grams × 2^(10/2.5) = 5 grams × 2^4 = 5 grams × 16 = 80 grams.

c)

To find the half-life of Element X in hours, we can use the formula:

Half-Life (in hours) = Time (in days) × 24 (to convert days to hours) ÷ log(Initial Mass / Final Mass)

Plugging in the values, we have Half-Life (in hours) = 1 day × 24 hours ÷ log(100 kg / 12.5 kg) = 1 day × 24 hours ÷ log(8) ≈ 1 day × 24 hours ÷ 0.9031 ≈ 1 day × 26.58 hours ≈ 26.58 hours.

d)

For a sample of Strontium-90 with a half-life of 28.1 years, if we want to know how many atoms remain undecayed after 56.2 years from a sample of 1.68 x 10^23 atoms, we can use the formula:

Remaining Atoms = Initial Atoms × (0.5)^(Time / Half-Life)

Plugging in the values, we have Remaining Atoms = 1.68 x 10^23 atoms × (0.5)^(56.2 years / 28.1 years) ≈ 1.68 x 10^23 atoms × 0.5^2 ≈ 1.68 x 10^23 atoms × 1/4 = 4.2 x 10^22 atoms.

e)

To find the percentage of a 900-gram sample of Potassium-37 that exists after 6.15 seconds with a half-life of 1.23 seconds, we can use the formula:

Percentage = (Final Mass / Initial Mass) × 100%

Plugging in the values, we have Percentage = (Final Mass / 900 grams) × 100% = (0.5)^(Time / Half-Life) × 100% = (0.5)^(6.15 seconds / 1.23 seconds) × 100% = (0.5)^5 × 100% ≈ 0.03125 × 100% = 3.125%.

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