From a decay curve for x parent isotope.

A. After 3 half-lives, x parental isotope, should have decayed by how many years?
B. How many milligrams of a 6.5 x 10^(-2)kg sample of the x parent isotope will reamin at the end of(A)?

1 answer

To answer your questions, we will need to use the concept of half-lives and how they affect the remaining amount of a radioactive isotope.

A. Time Decayed After 3 Half-Lives

The decay of a radioactive isotope is characterized by its half-life, which is the time required for half of the radioactive substance to decay.

Let’s denote the half-life of isotope X as \( t_{1/2} \). After 3 half-lives, the total time that has passed can be calculated as:

\[ \text{Total Time} = 3 \times t_{1/2} \]

To determine the actual duration in years, you would need to know \( t_{1/2} \) of the parent isotope X. Since this was not provided, I cannot give a specific numerical answer.

B. Remaining Amount After 3 Half-Lives

To find the amount of the isotope that remains after 3 half-lives, we can use the formula for remaining quantity based on half-lives:

\[ \text{Remaining Amount} = \text{Initial Amount} \times \left(\frac{1}{2}\right)^n \]

Where \( n \) is the number of half-lives passed. In this case, \( n = 3 \).

Given the initial mass of the parent isotope \( = 6.5 \times 10^{-2} \text{ kg} \), we can convert this into milligrams:

\[ 6.5 \times 10^{-2} \text{ kg} = 6.5 \times 10^{-2} \times 1000 \text{ mg} = 65 \text{ mg} \]

Now we can calculate the remaining amount after 3 half-lives:

\[ \text{Remaining Amount} = 65 \text{ mg} \times \left(\frac{1}{2}\right)^3 = 65 \text{ mg} \times \frac{1}{8} \]

\[ \text{Remaining Amount} = \frac{65}{8} \text{ mg} = 8.125 \text{ mg} \]

Summary

A. The total time after 3 half-lives is \( 3 \times t_{1/2} \) years (you would need the half-life to calculate the exact period).

B. 8.125 mg of the parent isotope will remain at the end of that time period.