In a radioactive substance at `t = 0`, the number of atoms is `8 xx 10^4`. Its half-life period is `3` years. The number of atoms `1 xx 10^4` will remain after interval.

Options

A. 6 years
B. 24 years
C. 3 years
D. 9 years

Answer

To solve the problem step by step, we will use the concepts of radioactive decay and the half-life of a substance.

### Step 1: Understand the given data
- Initial number of atoms, \( n_0 = 8 \times 10^4 \)
- Half-life period, \( t_{1/2} = 3 \) years
- Final number of atoms, \( n_t = 1 \times 10^4 \)

### Step 2: Write the formula for radioactive decay
The number of atoms remaining after time \( t \) can be expressed using the formula:
\[ n_t = n_0 \cdot e^{-\lambda t} \]
where \( \lambda \) is the decay constant.

### Step 3: Calculate the decay constant \( \lambda \)
The decay constant \( \lambda \) is related to the half-life by the formula:
\[ \lambda = \frac{\ln 2}{t_{1/2}} \]
Substituting the half-life:
\[ \lambda = \frac{\ln 2}{3} \]

### Step 4: Set up the equation with the known values
Substituting \( n_t \), \( n_0 \), and \( \lambda \) into the decay formula:
\[ 1 \times 10^4 = 8 \times 10^4 \cdot e^{-\left(\frac{\ln 2}{3}\right) t} \]

### Step 5: Simplify the equation
Dividing both sides by \( 8 \times 10^4 \):
\[ \frac{1 \times 10^4}{8 \times 10^4} = e^{-\left(\frac{\ln 2}{3}\right) t} \]
This simplifies to:
\[ \frac{1}{8} = e^{-\left(\frac{\ln 2}{3}\right) t} \]

### Step 6: Take the natural logarithm of both sides
Taking the natural logarithm:
\[ \ln\left(\frac{1}{8}\right) = -\left(\frac{\ln 2}{3}\right) t \]

### Step 7: Simplify the left side
The left side can be rewritten using properties of logarithms:
\[ \ln\left(\frac{1}{8}\right) = \ln(8^{-1}) = -\ln(8) \]
Thus, we have:
\[ -\ln(8) = -\left(\frac{\ln 2}{3}\right) t \]
Cancelling the negative signs gives:
\[ \ln(8) = \left(\frac{\ln 2}{3}\right) t \]

### Step 8: Express \( \ln(8) \) in terms of \( \ln(2) \)
Since \( 8 = 2^3 \):
\[ \ln(8) = \ln(2^3) = 3 \ln(2) \]
Substituting this back into the equation:
\[ 3 \ln(2) = \left(\frac{\ln 2}{3}\right) t \]

### Step 9: Solve for \( t \)
Now, dividing both sides by \( \ln(2) \):
\[ 3 = \frac{t}{3} \]
Multiplying both sides by 3 gives:
\[ t = 9 \text{ years} \]

### Final Answer
The time at which the number of atoms will remain \( 1 \times 10^4 \) is \( t = 9 \) years.

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