You are working with two isotopes of an element (i.e. the same element,
but with different masses and hence different half-lives). Isotope A has a half life
of 10 days, whereas Isotope B has a half life of 5 days.
At time t=0, you detect that the mixture experiences 1010 atoms decaying per
second. After 20 days, the decay activity has reduced to 20% of what it was at
time t=0.
How many atoms decay per second at time t=0 of both Isotope A and Isotope
B?


We are given two isotopes, A and B, with half-lives of 10 days and 5 days, respectively. At time \( t = 0 \), the total decay activity is \( 1010 \) atoms per second, and after 20 days, the decay activity is reduced to \( 20\% \) of the initial activity.

We need to determine the initial number of atoms decaying per second from each isotope at \( t = 0 \).

### Step 1: Use of the Decay Law

The decay law is given by:
\[
A(t) = A_0 e^{-\lambda t}
\]
where:
- \( A(t) \) is the activity at time \( t \),
- \( A_0 \) is the initial activity at time \( t = 0 \),
- \( \lambda \) is the decay constant, related to the half-life \( T \) by \( \lambda = \frac{\ln(2)}{T} \),
- \( t \) is the time elapsed.

For isotope A (with half-life \( T_A = 10 \) days):
\[
\lambda_A = \frac{\ln(2)}{10} \text{ days}^{-1}
\]

For isotope B (with half-life \( T_B = 5 \) days):
\[
\lambda_B = \frac{\ln(2)}{5} \text{ days}^{-1}
\]

### Step 2: Setting Up the Equations

Let the initial activity of isotope A be \( A_{0A} \) and that of isotope B be \( A_{0B} \). So, the total initial activity is:

\[
A_{0A} + A_{0B} = 1010 \text{ atoms/second} \tag{1}
\]

After 20 days, the total activity has reduced to \( 20\% \) of the initial activity, meaning:

\[
A(20) = 0.2 \times 1010 = 202 \text{ atoms/second}
\]

The total activity at \( t = 20 \) days is given by the sum of the individual activities of the isotopes:

\[
A(20) = A_{0A} e^{-\lambda_A \cdot 20} + A_{0B} e^{-\lambda_B \cdot 20} \tag{2}
\]

Substitute \( \lambda_A \) and \( \lambda_B \) into equation (2).

### Step 3: Substituting the Decay Constants

Using \( \lambda_A = \frac{\ln(2)}{10} \approx 0.0693 \, \text{days}^{-1} \) and \( \lambda_B = \frac{\ln(2)}{5} \approx 0.1386 \, \text{days}^{-1} \), we substitute into equation (2):

\[
202 = A_{0A} e^{-0.0693 \cdot 20} + A_{0B} e^{-0.1386 \cdot 20}
\]

Simplifying the exponents:

\[
202 = A_{0A} e^{-1.386} + A_{0B} e^{-2.772}
\]
\[
202 = A_{0A} \times 0.250 + A_{0B} \times 0.0625 \tag{3}
\]

### Step 4: Solve the System of Equations

Now we have the following system of equations:
1. \( A_{0A} + A_{0B} = 1010 \) (from equation (1))
2. \( 0.250 A_{0A} + 0.0625 A_{0B} = 202 \) (from equation (3))

We can solve this system to find \( A_{0A} \) and \( A_{0B} \).

Let me calculate these values.The initial activities of the two isotopes are as follows:

- The activity of Isotope A at \( t = 0 \) is approximately \( 740.67 \) atoms per second.
- The activity of Isotope B at \( t = 0 \) is approximately \( 269.33 \) atoms per second.

These values sum up to the total initial decay activity of 1010 atoms per second.

Would you like further clarification on any step?

Here are five related questions to deepen your understanding:
1. How does the half-life of an isotope affect its decay activity over time?
2. How would the system change if the half-lives of the isotopes were swapped?
3. Can the total number of atoms of each isotope be determined from this data?
4. How does the decay constant \( \lambda \) relate to the rate of decay in practical scenarios?
5. How would we model the decay activity if a third isotope was present in the mixture?

**Tip**: The decay constant \( \lambda \) is a useful parameter that simplifies modeling decay processes and can be derived directly from the half-life.

You are working with two isotopes of an element (i.e. the same element, but with different masses and hence different half-lives). Isotope A has a half life of 10 days, whereas Isotope B has a half life of 5 days. At time t=0, you detect that the mixture experiences 1010 atoms decaying per second. After 20 days, the decay activity has reduced to 20% of what it was at time t=0. How many atoms decay per second at time t=0 of both Isotope A and Isotope B?

Learn how to calculate the initial decay activity of two isotopes with different half-lives using the exponential decay law. The problem involves determining the initial atoms decaying per second for Isotope A (10-day half-life) and Isotope B (5-day half-life). After 20 days, the decay activity reduces to 20% of the initial value. This solution demonstrates key formulas and concepts, ideal for undergraduate students.

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