Question

A radioactive material decays by simultaneous emissions of two particles with half lives of 1400 years and 700 years respectively. What will be the time after which one third of the material remains ? (Take \(\ln 3 = 1.1\))

• A. 740 years
• B. 1110 years
• C. 700 years
• D. 340 years

Hint:

Decay constants add for simultaneous decay modes: \(\lambda = \lambda_1 + \lambda_2\). Then \(N = N_0e^{-\lambda t}\) with \(N = N_0/3\).

Answer 1

The Correct Option is A

\(\lambda_1 = \frac{\ln 2}{1400}\), \(\lambda_2 = \frac{\ln 2}{700}\), so \(\lambda = \lambda_1 + \lambda_2 = \frac{3\ln 2}{1400}\) per year.
For one third remaining: \(\frac{N_0}{3} = N_0e^{-\lambda t}\), so \(t = \frac{\ln 3}{\lambda}\).
\(t = \frac{1.1 \times 1400}{3 \times 0.693} = \frac{1540}{2.079} \approx 740\) years.
Answer: (1) 740 years