Question

A radio active material is reduced to 1/8 of its original amount in 3 days.
If 8×10^-3 kg of the material is left after 5 days the initial amount of the material is

• A. 32g
• B. 40g
• C. 256g
• D. 64g

Answer 1

The Correct Option is C

The decay of radioactive material is described by the equation: \[ m = m_0 e^{-\lambda t}, \] where: 
 \( m_0 \) is the initial amount of the material, 
 \( m \) is the amount remaining after time \( t \), 
 \( \lambda \) is the decay constant, \item \( t \) is the time. 
Step 1: Determine the decay constant. After 3 days, the material is reduced to \( \frac{1}{8} \) of its initial amount: \[ \frac{m_0}{8} = m_0 e^{-\lambda \cdot 3}. \] Divide through by \( m_0 \): \[ \frac{1}{8} = e^{-3\lambda}. \] Take the natural logarithm: \[ \ln \frac{1}{8} = -3\lambda. \] Simplify: \[ \lambda = \frac{\ln 8}{3}. \] 
Step 2: Relate the remaining material after 5 days. After 5 days, the remaining material is: \[ m = m_0 e^{-\lambda \cdot 5}. \] Substitute \( m = 8 \times 10^{-3} \, \text{kg} = 8 \, \text{g} \) and \( \lambda = \frac{\ln 8}{3} \): \[ 8 = m_0 e^{-\frac{\ln 8}{3} \cdot 5}. \] Simplify the exponent: \[ e^{-\frac{\ln 8}{3} \cdot 5} = e^{-\ln 8 \cdot \frac{5}{3}} = e^{-\ln 2^3 \cdot \frac{5}{3}} = e^{-\ln 2^5} = \frac{1}{32}. \] Substitute: \[ 8 = m_0 \cdot \frac{1}{32}. \] Solve for \( m_0 \): \[ m_0 = 8 \cdot 32 = 256 \, \text{g}. \] 
Final Answer: The initial amount of the material is: \[ \boxed{256 \, \text{g}}. \]