Question

A and B are two radioactive samples of half-life 12 hours and 16 hours respectively. The number of nuclei in them are in the ratio 2:1. After 48 hours, this ratio will become

• A. 1:1
• B. 2:1
• C. 1:2
• D. 1:4

Hint:

For radioactive decay problems, use the formula \( N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T}} \) to find the remaining number of nuclei after a certain time.

Answer 1

The Correct Option is A

Step 1: Understand the decay formula.
The number of nuclei remaining in a radioactive sample after time \( t \) is given by the equation:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T}} \] where: \( N_0 \) is the initial number of nuclei, \( t \) is the time elapsed, \( T \) is the half-life of the substance.
Step 2: Use the given values for A and B.

For sample A, \( T_A = 12 \) hours, For sample B, \( T_B = 16 \) hours, After 48 hours, we calculate the remaining number of nuclei in each sample using the decay formula.
For sample A: \[ N_A(48) = N_{A0} \left( \frac{1}{2} \right)^{\frac{48}{12}} = N_{A0} \left( \frac{1}{2} \right)^4 = N_{A0} \times \frac{1}{16} \] For sample B: \[ N_B(48) = N_{B0} \left( \frac{1}{2} \right)^{\frac{48}{16}} = N_{B0} \left( \frac{1}{2} \right)^3 = N_{B0} \times \frac{1}{8} \] 
Step 3: Conclusion.
After 48 hours, the ratio of \( N_A(48) \) to \( N_B(48) \) becomes: \[ \frac{N_A(48)}{N_B(48)} = \frac{N_{A0} \times \frac{1}{16}}{N_{B0} \times \frac{1}{8}} = \frac{1}{2} \] Thus, the ratio of the number of nuclei becomes 1:1. 
Conclusion:
The correct answer is (A) 1:1.