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The isotope 90^38Sr in a 0.500-g sample diminishes to 0.393 g in 10.0 years. Calculate the half-life of 90^38Sr.

a) 8.0 years
b) 10.0 years
c) 12.0 years
d) 15.0 years

1 Answer

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Final answer:

To calculate the half-life of strontium-90, given a reduction in mass over 10 years, we use the formula for exponential decay and solve for the half-life. The calculation shows the half-life is approximately 28.8 years, which does not match any of the given options, suggesting a possible typo in the question or options.

Step-by-step explanation:

The student is asking to calculate the half-life of the isotope strontium-90 (9038Sr), given that a 0.500-g sample diminishes to 0.393 g in 10.0 years. We can use the half-life formula and the concept of exponential decay to solve this.

We know that the amount of a substance remaining after a certain number of half-lives can be expressed as:


(Remaining mass) = (Initial mass) × (1/2)(Time elapsed / Half-life)

So in the case of strontium-90, we have:


0.393 g = 0.500 g × (1/2)(10.0 years / Half-life)

When we solve for the half-life, we get:


Half-life = 10.0 years / (log(0.393/0.500) / log(0.5))

Plugging these values into a calculator, we find that the half-life of 9038Sr is approximately 28.8 years.

However, the options provided do not include this value. It might be a case where the problem has a typo or the given options are incorrect.

Based on the provided options, none of them are the correct calculation of half-life for 9038Sr as per the question's data. In such a case, it is suggested to double-check the problem statement or consult additional sources for the correct half-life value of 9038Sr if the provided answer options must be used.

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