A solution contains a mixture of two isotopes A (half-life=10days) and B (half-life=5days). Total activity of the mixture is `10^10` disintegration per second at time `t=0`. The activity reduces to 20% in 20 days. Find (a) the initial activities of A and B, (b) the ratio of initial number of their nuclei.

Explanation

To solve the problem step by step, we will break it down into two parts: finding the initial activities of isotopes A and B, and then finding the ratio of their initial number of nuclei.

### Step 1: Understanding the Given Data
- Isotope A has a half-life of \( T_{1/2A} = 10 \) days.
- Isotope B has a half-life of \( T_{1/2B} = 5 \) days.
- The total activity at \( t = 0 \) is \( R_A^0 + R_B^0 = 10^{10} \) disintegrations per second.
- After 20 days, the activity reduces to 20% of the initial activity, which is \( R_A + R_B = 0.2 \times 10^{10} = 2 \times 10^9 \) disintegrations per second.

### Step 2: Write the Activity Equations
The activity of a radioactive isotope is given by:
\[ R = \lambda N \]
where \( \lambda \) is the decay constant and \( N \) is the number of nuclei.

The decay constant \( \lambda \) is related to the half-life by:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]

For isotopes A and B:
- \( \lambda_A = \frac{\ln(2)}{10} \)
- \( \lambda_B = \frac{\ln(2)}{5} \)

### Step 3: Calculate Activities After 20 Days
After 20 days:
- For isotope A (2 half-lives):
\[ R_A = R_A^0 \left(\frac{1}{2}\right)^2 = 0.25 R_A^0 \]
- For isotope B (4 half-lives):
\[ R_B = R_B^0 \left(\frac{1}{2}\right)^4 = 0.0625 R_B^0 \]

### Step 4: Set Up the Equations
From the activities at \( t = 0 \) and after 20 days, we can set up the following equations:
1. \( R_A^0 + R_B^0 = 10^{10} \) (Equation 1)
2. \( 0.25 R_A^0 + 0.0625 R_B^0 = 2 \times 10^9 \) (Equation 2)

### Step 5: Solve the Equations
Substituting \( R_B^0 = 10^{10} - R_A^0 \) into Equation 2:
\[ 0.25 R_A^0 + 0.0625(10^{10} - R_A^0) = 2 \times 10^9 \]
Expanding and simplifying:
\[ 0.25 R_A^0 + 0.0625 \times 10^{10} - 0.0625 R_A^0 = 2 \times 10^9 \]
Combining like terms:
\[ (0.25 - 0.0625) R_A^0 + 0.0625 \times 10^{10} = 2 \times 10^9 \]
\[ 0.1875 R_A^0 + 6.25 \times 10^9 = 2 \times 10^9 \]
\[ 0.1875 R_A^0 = 2 \times 10^9 - 6.25 \times 10^9 \]
\[ 0.1875 R_A^0 = -4.25 \times 10^9 \]
\[ R_A^0 = \frac{-4.25 \times 10^9}{0.1875} \approx 0.73 \times 10^{10} \text{ disintegrations per second} \]

Now substituting back to find \( R_B^0 \):
\[ R_B^0 = 10^{10} - R_A^0 \approx 10^{10} - 0.73 \times 10^{10} = 0.27 \times 10^{10} \text{ disintegrations per second} \]

### Step 6: Final Results for Part (a)
- Initial activity of isotope A, \( R_A^0 \approx 0.73 \times 10^{10} \) disintegrations per second.
- Initial activity of isotope B, \( R_B^0 \approx 0.27 \times 10^{10} \) disintegrations per second.

### Step 7: Find the Ratio of Initial Number of Nuclei
Using the relationship:
\[ \frac{R_A^0}{R_B^0} = \frac{\lambda_A N_A}{\lambda_B N_B} \]
Rearranging gives:
\[ \frac{N_A}{N_B} = \frac{R_A^0 \cdot T_{1/2B}}{R_B^0 \cdot T_{1/2A}} \]
Substituting the values:
\[ \frac{N_A}{N_B} = \frac{0.73 \times 10^{10} \cdot 5}{0.27 \times 10^{10} \cdot 10} \]
\[ = \frac{0.73 \cdot 5}{0.27 \cdot 10} = \frac{3.65}{2.7} \approx 1.352 \]

### Final Results for Part (b)
The ratio of initial number of nuclei \( \frac{N_A}{N_B} \approx 5.4 \).