A radioactive substance has a half-life of 10 hours. Calculate the activity of `10^(-10)` gram of the substance after 2 hours if its atomic mass is 102.

Answer

To solve the problem, we need to calculate the activity of a radioactive substance after a certain time has passed. Here's a step-by-step solution:

### Step 1: Understand the Given Data
- **Half-life (T_half)** = 10 hours
- **Initial mass (m_0)** = \(10^{-10}\) grams
- **Atomic mass (M)** = 102 g/mol
- **Time (t)** = 2 hours

### Step 2: Calculate the Decay Constant (λ)
The decay constant (λ) is given by the formula:
\[ \lambda = \frac{\ln(2)}{T_{half}} \]
Substituting the values:
\[ \lambda = \frac{0.693}{10 \text{ hours}} = 0.0693 \text{ hours}^{-1} \]

### Step 3: Calculate the Remaining Mass After 2 Hours
Using the formula for radioactive decay:
\[ m(t) = m_0 \cdot e^{-\lambda t} \]
Substituting the known values:
\[ m(2) = 10^{-10} \cdot e^{-0.0693 \cdot 2} \]
Calculating the exponent:
\[ e^{-0.0693 \cdot 2} \approx e^{-0.1386} \approx 0.870 \]
Thus,
\[ m(2) \approx 10^{-10} \cdot 0.870 \approx 8.70 \times 10^{-11} \text{ grams} \]

### Step 4: Calculate the Number of Atoms (N)
To find the number of atoms, we use the formula:
\[ N = \frac{m(t)}{M} \cdot N_A \]
Where \(N_A\) (Avogadro's number) = \(6.022 \times 10^{23} \text{ mol}^{-1}\).

Substituting the values:
\[ N = \frac{8.70 \times 10^{-11}}{102} \cdot 6.022 \times 10^{23} \]
Calculating:
\[ N \approx \frac{8.70 \times 10^{-11}}{102} \approx 8.52 \times 10^{-13} \text{ mol} \]
Then,
\[ N \approx 8.52 \times 10^{-13} \cdot 6.022 \times 10^{23} \approx 5.13 \times 10^{11} \text{ atoms} \]

### Step 5: Calculate the Activity (A)
The activity is given by:
\[ A = \lambda N \]
Substituting the values:
\[ A = 0.0693 \cdot 5.13 \times 10^{11} \]
Calculating:
\[ A \approx 3.55 \times 10^{10} \text{ disintegrations per second (Bq)} \]

### Final Answer
The activity of the substance after 2 hours is approximately \(3.55 \times 10^{10} \text{ Bq}\).

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