The count rate of a radioactive sample falls from `4.0xx10^6 s^(-1)` to `1.0xx10^6 s^(-1)` in 20 hours. What will be the count rate after 100 hours from beginning ?

Options

A. `3.91xx10^3 s^(-1)`
B. `3.91xx10^2 s^(-1)`
C. `3.91xx10^4 s^(-1)`
D. `3.91xx10^6 s^(-1)`

Explanation

To find the count rate of a radioactive sample after 100 hours, given that the count rate falls from \(4.0 \times 10^6 \, s^{-1}\) to \(1.0 \times 10^6 \, s^{-1}\) in 20 hours, we can follow these steps:

### Step 1: Determine the initial and final count rates
- Initial count rate, \( A_0 = 4.0 \times 10^6 \, s^{-1} \)
- Final count rate after 20 hours, \( A = 1.0 \times 10^6 \, s^{-1} \)

### Step 2: Calculate the fraction of the count rate that remains after 20 hours
The count rate falls from \( A_0 \) to \( A \):
\[ \text{Fraction remaining} = \frac{A}{A_0} = \frac{1.0 \times 10^6}{4.0 \times 10^6} = \frac{1}{4} \]

### Step 3: Relate the fraction remaining to half-lives
Since the count rate falls to \( \frac{1}{4} \) of its original value in 20 hours, we can deduce that this corresponds to 2 half-lives because:
- After 1 half-life, the count rate would be \( \frac{1}{2} A_0 \)
- After 2 half-lives, the count rate would be \( \frac{1}{4} A_0 \)

### Step 4: Calculate the half-life
Given that 20 hours corresponds to 2 half-lives:
\[ \text{Half-life} = \frac{20 \, \text{hours}}{2} = 10 \, \text{hours} \]

### Step 5: Determine the number of half-lives in 100 hours
To find the count rate after 100 hours, we need to calculate how many half-lives fit into 100 hours:
\[ \text{Number of half-lives} = \frac{100 \, \text{hours}}{10 \, \text{hours}} = 10 \]

### Step 6: Calculate the remaining count rate after 10 half-lives
The remaining count rate after \( n \) half-lives can be calculated using the formula:
\[ A_n = A_0 \left( \frac{1}{2} \right)^n \]
Substituting \( n = 10 \):
\[ A_{10} = 4.0 \times 10^6 \left( \frac{1}{2} \right)^{10} \]
Calculating \( \left( \frac{1}{2} \right)^{10} = \frac{1}{1024} \):
\[ A_{10} = 4.0 \times 10^6 \times \frac{1}{1024} \approx 3.91 \times 10^3 \, s^{-1} \]

### Step 7: Final answer
Thus, the count rate after 100 hours is approximately:
\[ A_{10} \approx 3.91 \times 10^3 \, s^{-1} \]