A radioactive substances decays so that 3% of its initial nuclei remain after 60 seconds. The half life period of the substances is nearly

Options

A. 17.3 s
B. 12 s
C. 30 s
D. 60 s

Answer

To solve the problem of finding the half-life period of a radioactive substance that has 3% of its initial nuclei remaining after 60 seconds, we can follow these steps:

### Step 1: Define the initial and remaining nuclei
Let the initial number of nuclei be \( N_0 \). After 60 seconds, the number of remaining nuclei is given as 3% of \( N_0 \):
\[ N_t = \frac{3}{100} N_0 = 0.03 N_0 \]

### Step 2: Use the decay formula
The decay of radioactive substances can be described by the equation:
\[ N_t = N_0 e^{-\lambda t} \]
where:
- \( N_t \) is the number of nuclei remaining after time \( t \),
- \( N_0 \) is the initial number of nuclei,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

Substituting the values we have:
\[ 0.03 N_0 = N_0 e^{-\lambda \cdot 60} \]

### Step 3: Simplify the equation
We can cancel \( N_0 \) from both sides (assuming \( N_0 \neq 0 \)):
\[ 0.03 = e^{-\lambda \cdot 60} \]

### Step 4: Take the natural logarithm
Taking the natural logarithm of both sides gives:
\[ \ln(0.03) = -\lambda \cdot 60 \]

### Step 5: Solve for the decay constant \( \lambda \)
Rearranging the equation to solve for \( \lambda \):
\[ \lambda = -\frac{\ln(0.03)}{60} \]

### Step 6: Calculate \( \lambda \)
Calculating \( \ln(0.03) \):
\[ \ln(0.03) \approx -3.5066 \]
Thus,
\[ \lambda = -\frac{-3.5066}{60} \approx 0.05844 \, \text{s}^{-1} \]

### Step 7: Use the half-life formula
The half-life \( t_{1/2} \) is related to the decay constant \( \lambda \) by the formula:
\[ t_{1/2} = \frac{0.693}{\lambda} \]

### Step 8: Calculate the half-life
Substituting the value of \( \lambda \):
\[ t_{1/2} = \frac{0.693}{0.05844} \approx 11.87 \, \text{s} \]

### Step 9: Round the answer
Rounding this value gives:
\[ t_{1/2} \approx 12 \, \text{s} \]

### Final Answer
The half-life period of the substance is approximately **12 seconds**.

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