Radioactive Decay & Half-Life
Apply first-order kinetics to compare the decay rates of two simultaneous processes.
Question:
$A$ and $B$ are two substances undergoing radioactive decay in a container.
The half life of $A$ is $15 \text{ min}$ and that of $B$ is $5 \text{ min}$. If the initial concentration of $B$ is $4$ times that of $A$ and they both start decaying at the same time, how much time will it take for the concentration of both of them to be same?
Time taken = ________ $\text{min}$.
Detailed Solution
Step 1: Set up the given equations
Let the initial concentration of $A$ be $N_0$. Therefore, initial concentration of $B$ is $4N_0$.
Let the time taken for their concentrations to become equal be $t$.
- Concentration of $A$ at time $t$: $\quad N_A = \frac{N_0}{2^{t/15}}$
- Concentration of $B$ at time $t$: $\quad N_B = \frac{4N_0}{2^{t/5}}$
Step 2: Equate and Solve
We are given that at time $t$, $N_A = N_B$.
Cancel $N_0$ from both sides and cross-multiply:
$$ 2^{\left(\frac{t}{5} - \frac{t}{15}\right)} = 2^2 $$
Since the bases are the same, equate the exponents:
The Logic Shortcut (No Algebra!)
Don't want to mess with exponents? Use a simple mental table based on half-lives. We know $B$ decays much faster ($t_{1/2} = 5$ min). Let's trace it every $5$ minutes:
| Time ($t$) | Conc. of $B$ (Halves every 5m) | Conc. of $A$ ($t_{1/2} = 15$m) |
|---|---|---|
| $0 \text{ min}$ | $4N_0$ | $1N_0$ |
| $5 \text{ min}$ | $2N_0$ | Wait... |
| $10 \text{ min}$ | $1N_0$ | Wait... |
| $15 \text{ min}$ | $0.5N_0$ | $0.5N_0$ (1 Half-life) |
At 15 minutes, $B$ has undergone exactly 3 half-lives ($4 \to 2 \to 1 \to 0.5$), and $A$ has undergone exactly 1 half-life ($1 \to 0.5$). They match perfectly!
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