The initial count rate for a radioactive source was found to be 1600 count/sec.and t = 8 sec this rate was 100 count/sec. What was the count rate at t = 6 [in counts/sec]-

Options

A. `4000`
B. `300`
C. `200`
D. `150`

Answer

To solve the problem, we will use the radioactive decay formula. The steps are as follows:

### Step 1: Understand the decay formula
The decay of a radioactive substance can be described by the equation:
\[ \frac{n}{n_0} = e^{-\lambda t} \]
where:
- \( n_0 \) = initial count rate
- \( n \) = count rate at time \( t \)
- \( \lambda \) = decay constant
- \( t \) = time

### Step 2: Identify the known values
From the problem, we have:
- Initial count rate \( n_0 = 1600 \) counts/sec
- Count rate at \( t = 8 \) sec is \( n = 100 \) counts/sec
- We need to find \( n \) at \( t = 6 \) sec.

### Step 3: Use the decay formula to find \( \lambda \)
Using the values at \( t = 8 \) sec:
\[ \frac{100}{1600} = e^{-\lambda \cdot 8} \]
This simplifies to:
\[ \frac{1}{16} = e^{-8\lambda} \]

### Step 4: Take the natural logarithm of both sides
Taking the natural logarithm:
\[ \ln\left(\frac{1}{16}\right) = -8\lambda \]
This can be rewritten as:
\[ -4\ln(2) = -8\lambda \]
Thus, solving for \( \lambda \):
\[ \lambda = \frac{\ln(2)}{2} \]

### Step 5: Substitute \( \lambda \) back into the decay formula for \( t = 6 \) sec
Now we need to find \( n \) at \( t = 6 \) sec:
\[ \frac{n}{1600} = e^{-\lambda \cdot 6} \]
Substituting \( \lambda \):
\[ \frac{n}{1600} = e^{-\left(\frac{\ln(2)}{2}\right) \cdot 6} \]
This simplifies to:
\[ \frac{n}{1600} = e^{-3\ln(2)} = \frac{1}{2^3} = \frac{1}{8} \]

### Step 6: Solve for \( n \)
Now, multiplying both sides by 1600:
\[ n = 1600 \cdot \frac{1}{8} = 200 \text{ counts/sec} \]

### Final Answer
The count rate at \( t = 6 \) sec is \( 200 \) counts/sec.

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