A mixture consists of two radioactive materials $A_1$ and $A_2$ with half lives of 20 s and 10 s respectively.

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:

Accepted Answer

1. **Formula:** Radioactive decay follows first-order kinetics. The amount of substance remaining ($N$) after time $t$ is given by $N = N_0 (1/2)^n$, where $n = t/T_{1/2}$ is the number of half-lives .
2. **Setup Equation:** We want the time $t$ when the remaining amounts of $A_1$ and $A_2$ are equal ($N_1 = N_2$).
   $$ 40 \left(\frac{1}{2}ight)^{\frac{t}{20}} = 160 \left(\frac{1}{2}ight)^{\frac{t}{10}} $$
3. **Solve for t:**
   $$ \frac{(1/2)^{t/20}}{(1/2)^{t/10}} = \frac{160}{40} $$
   $$ \left(\frac{1}{2}ight)^{\frac{t}{20} - \frac{t}{10}} = 4 $$
   $$ \left(\frac{1}{2}ight)^{-\frac{t}{20}} = 2^2 $$
   $$ 2^{\frac{t}{20}} = 2^2 $$
   $$ \frac{t}{20} = 2 \implies t = 40 	ext{ s} $$
4. **Verification:** At $t=40$ s:
   - $A_1$ (2 half-lives): $40 	o 20 	o 10$ g.
   - $A_2$ (4 half-lives): $160 	o 80 	o 40 	o 20 	o 10$ g.
   - Amounts are equal.

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