A radioactive substance has a half-life of 64.8 h. A sample containing this isotope has an initial activity `(t=0)` of `40muCi`. Calculate the number of nuclei that decay in the time interval between `t_1=10.0h` and `t_2=12.0h`.

Explanation

To solve the problem of calculating the number of nuclei that decay between the time intervals \( t_1 = 10.0 \, \text{h} \) and \( t_2 = 12.0 \, \text{h} \) for a radioactive substance with a half-life of \( 64.8 \, \text{h} \) and an initial activity of \( 40 \, \mu\text{Ci} \), we can follow these steps:

### Step 1: Calculate the Decay Constant (\( \lambda \))

The decay constant \( \lambda \) can be calculated using the formula:

\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]

where \( T_{1/2} \) is the half-life.

Given:
- \( T_{1/2} = 64.8 \, \text{h} \)

Calculating \( \lambda \):

\[ \lambda = \frac{0.693}{64.8} \approx 0.0107 \, \text{h}^{-1} \]

### Step 2: Calculate the Initial Number of Nuclei (\( N_0 \))

The initial activity \( A_0 \) is related to the decay constant and the initial number of nuclei \( N_0 \) by the formula:

\[ A_0 = \lambda N_0 \]

Rearranging gives:

\[ N_0 = \frac{A_0}{\lambda} \]

Given:
- \( A_0 = 40 \, \mu\text{Ci} = 40 \times 10^{-6} \, \text{Ci} \)
- \( 1 \, \text{Ci} = 3.7 \times 10^{10} \, \text{decays/s} \)

Calculating \( A_0 \) in decays per hour:

\[ A_0 = 40 \times 10^{-6} \times 3.7 \times 10^{10} \, \text{decays/h} \approx 1.48 \times 10^6 \, \text{decays/h} \]

Now substituting into the equation for \( N_0 \):

\[ N_0 = \frac{1.48 \times 10^6}{0.0107} \approx 1.38 \times 10^8 \]

### Step 3: Calculate the Number of Nuclei at \( t_1 \) and \( t_2 \)

Using the formula for radioactive decay:

\[ N(t) = N_0 e^{-\lambda t} \]

**At \( t_1 = 10.0 \, \text{h} \)**:

\[ N_1 = N_0 e^{-\lambda t_1} = 1.38 \times 10^8 e^{-0.0107 \times 10} \approx 1.24 \times 10^8 \]

**At \( t_2 = 12.0 \, \text{h} \)**:

\[ N_2 = N_0 e^{-\lambda t_2} = 1.38 \times 10^8 e^{-0.0107 \times 12} \approx 1.214 \times 10^8 \]

### Step 4: Calculate the Number of Decayed Nuclei

The number of nuclei that decay between \( t_1 \) and \( t_2 \) is given by:

\[ \Delta N = N_1 - N_2 \]

Calculating:

\[ \Delta N = 1.24 \times 10^8 - 1.214 \times 10^8 \approx 0.026 \times 10^8 \approx 2.6 \times 10^6 \]

### Final Answer

The number of nuclei that decay between \( t_1 = 10.0 \, \text{h} \) and \( t_2 = 12.0 \, \text{h} \) is approximately \( 2.6 \times 10^6 \).

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