The counting rate observed from a radioactive source at t=0 second was 1600 counts per second and t =8 seconds it was 100 counts per second. The counting rate observed as counts per second t = 6 seconds will be

Options

A. 250
B. 300
C. 250
D. 200

Answer

To solve the problem step-by-step, we will use the decay formula for radioactive materials. The formula is given by:

\[ A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]

where:
- \( A \) is the remaining activity (count rate) at time \( t \),
- \( A_0 \) is the initial activity (count rate) at time \( t = 0 \),
- \( T_{1/2} \) is the half-life of the radioactive material,
- \( t \) is the elapsed time.

### Step 1: Identify the known values
From the problem statement:
- At \( t = 0 \) seconds, \( A_0 = 1600 \) counts per second.
- At \( t = 8 \) seconds, \( A = 100 \) counts per second.

### Step 2: Use the decay formula to find the half-life
Substituting the known values into the decay formula:

\[ 100 = 1600 \left(\frac{1}{2}\right)^{\frac{8}{T_{1/2}}} \]

### Step 3: Simplify the equation
Dividing both sides by 1600:

\[ \frac{100}{1600} = \left(\frac{1}{2}\right)^{\frac{8}{T_{1/2}}} \]

This simplifies to:

\[ \frac{1}{16} = \left(\frac{1}{2}\right)^{\frac{8}{T_{1/2}}} \]

### Step 4: Rewrite \( \frac{1}{16} \) as a power of \( \frac{1}{2} \)
We know that:

\[ \frac{1}{16} = \left(\frac{1}{2}\right)^{4} \]

So we can equate the exponents:

\[ 4 = \frac{8}{T_{1/2}} \]

### Step 5: Solve for the half-life \( T_{1/2} \)
Rearranging gives:

\[ T_{1/2} = \frac{8}{4} = 2 \text{ seconds} \]

### Step 6: Find the count rate at \( t = 6 \) seconds
Now we will use the half-life to find the count rate at \( t = 6 \) seconds. Using the decay formula again:

\[ A = 1600 \left(\frac{1}{2}\right)^{\frac{6}{2}} \]

### Step 7: Simplify the expression
Calculating the exponent:

\[ A = 1600 \left(\frac{1}{2}\right)^{3} \]

This simplifies to:

\[ A = 1600 \cdot \frac{1}{8} \]

### Step 8: Calculate the final count rate
Calculating gives:

\[ A = 200 \text{ counts per second} \]

Thus, the count rate observed at \( t = 6 \) seconds is **200 counts per second**.

### Final Answer
The counting rate observed at \( t = 6 \) seconds is **200 counts per second**.

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