What is the half-life of an isotope that decays to 3.125% of its original activity in 50.4 h?

How can we calculate the half-life of an isotope that decays to 3.125% of its original activity in 50.4 hours?

Given the data, what formula and steps can we use to determine the half-life of the isotope?

Calculating the Half-Life of the Isotope

In order to calculate the half-life of an isotope that decays to 3.125% of its original activity in 50.4 hours, we need to use the following formula and steps:

First, we know that the half-life formula is t1/2 = (ln 2) / λ, where λ is the decay constant. Rearranging the formula gives us λ = (ln 2) / t1/2.

Next, we can substitute the given values into the formula. The isotope decays to 3.125% of its original activity in 50.4 hours, which means its activity at that point is 0.03125.

By substituting the values Ao = 1, t = 50.4 hours, and Activity = 0.03125 into the formula Activity = Ao * e^(−λt), we can find the value of λ, which is approximately -0.0137.

After finding the value of λ, we can then substitute it back into the half-life formula t1/2 = 0.693 / λ to calculate the half-life of the isotope, which is approximately 50.6 hours.

However, since the activity in question decreases to 3.125% of its original value (1/32 of the original value), it would require 5 half-lives for the activity to reach that level. Therefore, the time it takes for the activity to decrease to 3.125% of its original value is 253 hours (approximately 10.5 days) or 6072 hours (approximately 288 days).

By following these calculations, we can determine the half-life of the isotope based on the given data.