The half-life of a radioactive substance against `alpha-`decay is `1.2 xx 10^7 s`. What is the decay rate for `4 xx 10^15` atoms of the substance ?

Options

A. `2.3 xx 10^8` atom/s
B. `3.2 xx 10^8` atom/s
C. `2.3 xx10^(11)` atom/s
D. `3.2 xx 10^(11)` atom/s

Answer

To solve the problem, we need to find the decay rate (activity) of a radioactive substance given its half-life and the initial number of atoms. Here’s a step-by-step solution:

### Step 1: Understand the relationship between half-life and decay constant (λ)
The half-life (T₁/₂) of a radioactive substance is related to the decay constant (λ) by the formula:
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]
From this, we can express the decay constant as:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]

### Step 2: Substitute the given half-life into the equation
Given that the half-life \(T_{1/2} = 1.2 \times 10^7 \, \text{s}\), we can calculate λ:
\[ \lambda = \frac{\ln(2)}{1.2 \times 10^7} \]
Using the value of \(\ln(2) \approx 0.693\):
\[ \lambda \approx \frac{0.693}{1.2 \times 10^7} \approx 5.775 \times 10^{-8} \, \text{s}^{-1} \]

### Step 3: Calculate the decay rate (activity)
The decay rate (activity, A) is given by the formula:
\[ A = \lambda \cdot N \]
where \(N\) is the number of radioactive atoms. Given \(N = 4 \times 10^{15}\) atoms, we can substitute the values:
\[ A = (5.775 \times 10^{-8} \, \text{s}^{-1}) \cdot (4 \times 10^{15}) \]

### Step 4: Perform the multiplication
Calculating the activity:
\[ A = 5.775 \times 10^{-8} \times 4 \times 10^{15} = 2.31 \times 10^{8} \, \text{atoms/s} \]

### Step 5: Round the answer
Rounding the answer gives:
\[ A \approx 2.3 \times 10^{8} \, \text{atoms/s} \]

### Final Answer
The decay rate for \(4 \times 10^{15}\) atoms of the substance is approximately \(2.3 \times 10^{8} \, \text{atoms/s}\).

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