Question

The half-lives of two radioactive materials A and B are respectively \( T \) and \( 2T \). If the ratio of the initial masses of the materials A and B is 8:1, then the time after which the ratio of the masses of the materials A and B becomes 4:1 is:

• A. \( 2T \)
• B. \( T \)
• C. \( 4T \)
• D. \( 8T \)

Hint:

Use the radioactive decay formula \( m = m_0 \left( \frac{1}{2} \right)^{\frac{t}{T}} \) for solving such problems.
- Take logarithm on both sides when solving for \( t \).

Answer 1

The Correct Option is A

The decay formula for a radioactive substance is: \[ m = m_0 \left( \frac{1}{2} \right)^{\frac{t}{T}} \] where: - \( m_0 \) is the initial mass, - \( m \) is the remaining mass after time \( t \), - \( T \) is the half-life of the substance. 1. Mass of A and B after time \( t \): \[ m_A = 8 \left( \frac{1}{2} \right)^{\frac{t}{T}} \] \[ m_B = 1 \left( \frac{1}{2} \right)^{\frac{t}{2T}} \] 2. Condition for given ratio: \[ \frac{m_A}{m_B} = 4 \] \[ \frac{8 \left( \frac{1}{2} \right)^{\frac{t}{T}}}{\left( \frac{1}{2} \right)^{\frac{t}{2T}}} = 4 \] 3. Solving for \( t \): \[ 8 \times \left( \frac{1}{2} \right)^{\frac{t}{T} - \frac{t}{2T}} = 4 \] \[ 8 \times \left( \frac{1}{2} \right)^{\frac{t}{2T}} = 4 \] Taking logarithm, \[ \frac{t}{2T} = 1 \] \[ t = 2T \] Thus, the correct answer is \(\boxed{2T}\).