The half lives of radiosotopes P32 are 14 days and 25 days respectively.The radioisotopes are mixed in the ratio of 4:1 of their atoms .If the initial activity of the mixed sample is 3.0m Ci.Find the activity of mixed isotopes after 60year.

To determine the activity of a mixed sample of radioisotopes after a certain period, we need to consider the half-lives of each isotope, their initial activity, and how they decay over time. In this case, we have two isotopes, P-32, with half-lives of 14 days and 25 days, mixed in a ratio of 4:1. Let's break this down step by step.

Understanding Initial Conditions

The initial activity of the mixed sample is given as 3.0 mCi. Since the isotopes are mixed in a ratio of 4:1, we can calculate the initial activities of each isotope.

• Let the activity of the first isotope (with a half-life of 14 days) be 4x.
• Let the activity of the second isotope (with a half-life of 25 days) be x.

From the ratio, we can express the total initial activity as:

4x + x = 3.0 mCi

5x = 3.0 mCi

Thus, x = 0.6 mCi.

Now we can find the initial activities:

• Activity of the first isotope = 4x = 4 * 0.6 mCi = 2.4 mCi
• Activity of the second isotope = x = 0.6 mCi

Calculating Activity After 60 Years

Next, we need to calculate the activity of each isotope after 60 years. First, we convert 60 years into days:

60 years = 60 * 365 = 21,900 days.

Decay Formula

The activity of a radioactive substance after a certain time can be calculated using the formula:

A = A₀ * (1/2)^(t/T₁/2)

Where:

• A = remaining activity after time t
• A₀ = initial activity
• t = total time elapsed
• T₁/2 = half-life of the isotope

Activity of the First Isotope (14 days half-life)

For the first isotope:

• A₀ = 2.4 mCi
• T₁/2 = 14 days
• t = 21,900 days

Now, we calculate the number of half-lives that have passed:

Number of half-lives = t / T₁/2 = 21,900 days / 14 days ≈ 1564.29

Now we can find the remaining activity:

A₁ = 2.4 mCi * (1/2)^(1564.29) ≈ 0 mCi (essentially zero due to extensive decay)

Activity of the Second Isotope (25 days half-life)

For the second isotope:

• A₀ = 0.6 mCi
• T₁/2 = 25 days

Calculating the number of half-lives for the second isotope:

Number of half-lives = 21,900 days / 25 days ≈ 876

Now we find the remaining activity:

A₂ = 0.6 mCi * (1/2)^(876) ≈ 0 mCi (again, essentially zero)

Final Activity of the Mixed Sample

After 60 years, both isotopes have decayed to a point where their activities are negligible. Therefore, the total activity of the mixed sample after 60 years is:

A_total = A₁ + A₂ ≈ 0 mCi + 0 mCi = 0 mCi.

In summary, after 60 years, the activity of the mixed sample of P-32 isotopes will be virtually zero due to the extensive decay of both isotopes. This illustrates the significant impact of half-lives on the stability and activity of radioactive materials over long periods.