Two radioactive nuclides A and B have half-lives 50 min and 10 min respectively . A fresh sample contains the nuclides of B to be eight times that of A. How much time should elapse so that the mumber of nuclides of A becomes double of B ?

Options

A. 30
B. 40
C. 50
D. 100

Answer

To solve the problem, we will use the concept of radioactive decay and the half-life of the nuclides.

### Step-by-Step Solution:

1. **Define Initial Conditions**:
- Let the initial number of nuclides of A be \( N_A(0) \).
- Since the number of nuclides of B is eight times that of A, we have \( N_B(0) = 8N_A(0) \).

2. **Half-lives**:
- The half-life of nuclide A is \( t_{1/2}(A) = 50 \) minutes.
- The half-life of nuclide B is \( t_{1/2}(B) = 10 \) minutes.

3. **Decay Formula**:
The number of nuclides remaining after time \( t \) can be calculated using the formula:
\[ N(t) = N(0) \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]

4. **Calculate Remaining Nuclides**:
- For nuclide A after time \( t \):
\[ N_A(t) = N_A(0) \left( \frac{1}{2} \right)^{\frac{t}{50}} \]
- For nuclide B after time \( t \):
\[ N_B(t) = 8N_A(0) \left( \frac{1}{2} \right)^{\frac{t}{10}} \]

5. **Set Up the Equation**:
We want to find the time \( t \) when the number of nuclides of A becomes double that of B:
\[ N_A(t) = 2N_B(t) \]
Substituting the expressions for \( N_A(t) \) and \( N_B(t) \):
\[ N_A(0) \left( \frac{1}{2} \right)^{\frac{t}{50}} = 2 \left( 8N_A(0) \left( \frac{1}{2} \right)^{\frac{t}{10}} \right) \]

6. **Simplify the Equation**:
Cancel \( N_A(0) \) from both sides (assuming \( N_A(0) \neq 0 \)):
\[ \left( \frac{1}{2} \right)^{\frac{t}{50}} = 16 \left( \frac{1}{2} \right)^{\frac{t}{10}} \]
This can be rewritten as:
\[ \left( \frac{1}{2} \right)^{\frac{t}{50}} = \left( \frac{1}{2} \right)^{4} \left( \frac{1}{2} \right)^{\frac{t}{10}} \]
Combining the exponents:
\[ \left( \frac{1}{2} \right)^{\frac{t}{50}} = \left( \frac{1}{2} \right)^{4 + \frac{t}{10}} \]

7. **Equate the Exponents**:
Since the bases are the same, we can equate the exponents:
\[ \frac{t}{50} = 4 + \frac{t}{10} \]

8. **Solve for \( t \)**:
Multiply through by 50 to eliminate the fraction:
\[ t = 200 + 5t \]
Rearranging gives:
\[ t - 5t = 200 \implies -4t = 200 \implies t = -50 \]
This means \( t = 50 \) minutes.

### Final Answer:
The time that should elapse so that the number of nuclides of A becomes double that of B is **50 minutes**.