`._11^24Na`(half-life=15hrs.) is known to contain some radioactive impurity (half-life=3hrs.) in a sample. This sample has an intial activity of 1000 counts per minute, and after 30 hrs it shows an activity of 200 counts per minute. what percent of the intial activity was due to the impurity ?

Options

A. 10
B. 40
C. 5
D. 20

Explanation

To solve the problem, we need to analyze the decay of both the main isotope \( ^{24}_{11}Na \) and the radioactive impurity. Let's break down the steps:

### Step 1: Define Variables
Let:
- \( A \) = initial activity due to the impurity (in counts per minute)
- \( 1000 - A \) = initial activity due to \( ^{24}_{11}Na \)

### Step 2: Calculate the Decay of \( ^{24}_{11}Na \)
The half-life of \( ^{24}_{11}Na \) is 15 hours. After 30 hours (which is 2 half-lives), the activity will decay as follows:

\[ \text{Remaining activity of } ^{24}_{11}Na = (1000 - A) \times \left(\frac{1}{2}\right)^2 = (1000 - A) \times \frac{1}{4} \]

### Step 3: Calculate the Decay of the Impurity
The half-life of the impurity is 3 hours. After 30 hours (which is 10 half-lives), the activity will decay as follows:

\[ \text{Remaining activity of impurity} = A \times \left(\frac{1}{2}\right)^{10} = A \times \frac{1}{1024} \]

### Step 4: Set Up the Total Activity Equation
After 30 hours, the total activity is given as 200 counts per minute. Therefore, we can write the equation:

\[ \frac{1}{4}(1000 - A) + \frac{A}{1024} = 200 \]

### Step 5: Solve the Equation
Multiply through by 1024 to eliminate the fractions:

\[ 256(1000 - A) + A = 204800 \]

Expanding the equation gives:

\[ 256000 - 256A + A = 204800 \]

Combine like terms:

\[ 256000 - 204800 = 256A - A \]

\[ 51200 = 255A \]

Now, solve for \( A \):

\[ A = \frac{51200}{255} \approx 200.78 \text{ counts per minute} \]

### Step 6: Calculate the Percentage of Initial Activity Due to Impurity
To find the percentage of the initial activity that was due to the impurity, we use:

\[ \text{Percentage} = \left(\frac{A}{1000}\right) \times 100 \]

Substituting the value of \( A \):

\[ \text{Percentage} \approx \left(\frac{200.78}{1000}\right) \times 100 \approx 20.08\% \]

Thus, we can round this to approximately 20%.

### Final Answer
The percentage of the initial activity that was due to the impurity is approximately **20%**.

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