Question

A mixture consists of two radioactive materials \( A_1 \) and \( A_2 \) with half-lives of 20 s and 10 s respectively. Initially, the mixture has 40 g of \( A_1 \) and 160 g of \( A_2 \). The amount of the two in the mixture will become equal after

• A. 60 s
• B. 80 s
• C. 20 s
• D. 40 s

Hint:

Radioactive decay follows an exponential law, and solving for equal amounts requires equating the decay equations.

Answer 1

The Correct Option is D

The decay equation for a radioactive substance is: \[ N = N_0 \left(\frac{1}{2}\right)^{t/T} \] For \( A_1 \): \[ N_1 = 40 \left(\frac{1}{2}\right)^{t/20} \] For \( A_2 \): \[ N_2 = 160 \left(\frac{1}{2}\right)^{t/10} \] Setting \( N_1 = N_2 \): \[ 40 \left(\frac{1}{2}\right)^{t/20} = 160 \left(\frac{1}{2}\right)^{t/10} \] Solving for \( t \): \[ t = 40 s \] Thus, the correct answer is 40 s.