The count rate observed from a radioactive source at t sound was `N_0` and at 4t second it was `(N_0)/(16)`. The count rate observed at `(11/2)t` second will be

Options

A. (a) `(N_0)/(128)`
B. (b) `(N_0)/(64)`
C. (c) `(N_0)/(32)`
D. (d) None of these

Explanation

To solve the problem step by step, we will use the concept of half-life in radioactive decay.

### Step 1: Understand the given data
At time \( t \), the count rate is \( N_0 \). At time \( 4t \), the count rate is \( \frac{N_0}{16} \).

### Step 2: Relate count rates to half-lives
The decay of a radioactive substance can be described by the formula:
\[ N = N_0 \left( \frac{1}{2} \right)^n \]
where \( N \) is the count rate at time \( t \), \( N_0 \) is the initial count rate, and \( n \) is the number of half-lives that have passed.

### Step 3: Determine the number of half-lives
From the information given:
\[ \frac{N_0}{16} = N_0 \left( \frac{1}{2} \right)^n \]
Dividing both sides by \( N_0 \):
\[ \frac{1}{16} = \left( \frac{1}{2} \right)^n \]
Since \( \frac{1}{16} = \left( \frac{1}{2} \right)^4 \), we can equate the exponents:
\[ n = 4 \]
This means that 4 half-lives have passed from time \( t \) to \( 4t \).

### Step 4: Calculate the half-life
The time taken for 4 half-lives is:
\[ 4 \times \text{(half-life)} = 4t - t = 3t \]
Thus, the half-life \( T_{1/2} \) can be calculated as:
\[ 4 \times T_{1/2} = 3t \implies T_{1/2} = \frac{3t}{4} \]

### Step 5: Find the count rate at \( \frac{11}{2}t \)
Now we need to find the count rate at time \( \frac{11}{2}t \). The time elapsed from \( t \) to \( \frac{11}{2}t \) is:
\[ \frac{11}{2}t - t = \frac{9}{2}t \]
Now, we need to determine how many half-lives fit into \( \frac{9}{2}t \):
\[ \text{Number of half-lives} = \frac{\frac{9}{2}t}{T_{1/2}} = \frac{\frac{9}{2}t}{\frac{3}{4}t} = \frac{9}{2} \times \frac{4}{3} = 6 \]
So, 6 half-lives have passed.

### Step 6: Calculate the count rate after 6 half-lives
Using the decay formula:
\[ N = N_0 \left( \frac{1}{2} \right)^6 \]
Calculating \( \left( \frac{1}{2} \right)^6 \):
\[ \left( \frac{1}{2} \right)^6 = \frac{1}{64} \]
Thus, the count rate at \( \frac{11}{2}t \) is:
\[ N = N_0 \cdot \frac{1}{64} = \frac{N_0}{64} \]

### Final Answer
The count rate observed at \( \frac{11}{2}t \) seconds will be \( \frac{N_0}{64} \).

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