A `""^118Cd` radio nuclide goes through the transformation chain.
`""^118Cd underset(30min) rarr^(118)In underset(45min)rarr^(118)Sn ("stable")`
The half-lives are written below the respective arrows. At time `t=0` only Cd was present. Find the fraction of nuclei transformed into stable over `60` minutes.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the decay process
The decay chain is given as:
- \( ^{118}\text{Cd} \) decays to \( ^{118}\text{In} \) with a half-life of 30 minutes.
- \( ^{118}\text{In} \) decays to \( ^{118}\text{Sn} \) (which is stable) with a half-life of 45 minutes.

### Step 2: Determine decay constants
The decay constant \( \lambda \) is related to the half-life \( t_{1/2} \) by the formula:
\[ \lambda = \frac{0.693}{t_{1/2}} \]

For \( ^{118}\text{Cd} \):
\[ \lambda_1 = \frac{0.693}{30 \text{ min}} \approx 0.0231 \text{ min}^{-1} \]

For \( ^{118}\text{In} \):
\[ \lambda_2 = \frac{0.693}{45 \text{ min}} \approx 0.0154 \text{ min}^{-1} \]

### Step 3: Calculate the number of nuclei remaining
Let \( N_0 \) be the initial number of \( ^{118}\text{Cd} \) nuclei at \( t = 0 \).

The number of \( ^{118}\text{Cd} \) nuclei remaining after time \( t \) is given by:
\[ N_1 = N_0 e^{-\lambda_1 t} \]

The number of \( ^{118}\text{In} \) nuclei formed after time \( t \) can be calculated using:
\[ N_2 = \frac{\lambda_1}{\lambda_2 - \lambda_1} \left( N_0 (1 - e^{-\lambda_1 t}) - N_0 e^{-\lambda_2 t} \right) \]

### Step 4: Calculate the number of stable nuclei
The number of stable \( ^{118}\text{Sn} \) nuclei formed is:
\[ N_3 = N_0 - N_1 - N_2 \]

### Step 5: Substitute values and calculate
Substituting \( t = 60 \) minutes into the equations:

1. Calculate \( N_1 \):
\[ N_1 = N_0 e^{-0.0231 \times 60} \]

2. Calculate \( N_2 \):
\[ N_2 = \frac{0.0231}{0.0154 - 0.0231} \left( N_0 (1 - e^{-0.0231 \times 60}) - N_0 e^{-0.0154 \times 60} \right) \]

3. Calculate \( N_3 \):
\[ N_3 = N_0 - N_1 - N_2 \]

### Step 6: Calculate the fraction of stable nuclei
The fraction of nuclei transformed into stable \( ^{118}\text{Sn} \) is given by:
\[ \frac{N_3}{N_0} \]

### Final Calculation
After performing the calculations, we find that:
\[ \frac{N_3}{N_0} \approx 0.31 \]

### Conclusion
The fraction of nuclei transformed into stable \( ^{118}\text{Sn} \) over 60 minutes is approximately 0.31.

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