Two isotopes 'P' and 'Q' of atomic masses 10 and 20, respectively, are mixed in equal amount, by mas — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Track Decay of Isotope P Let initial mass of each isotope = $w_0$. Isotope P: $t_{1/2} = 10$ days. After $t = 20$ days: $$n_P = \frac{20}{10} = 2 \text{ half-lives}$$ $$w_P = w_0 \left(\frac{1}{2}\right)^2 = \frac{w_0}{4}$$ Step 2 - Use Mass Ratio to Find Remaining Mass of Q Given final ratio $w_P : w_Q = 1:4$: $$\frac{w_P}{w_Q} = \frac{1}{4} \implies w_Q = 4 w_P = 4 \cdot \frac{w_0}{4} = w_0$$ Step 3 - Deduce Half-Life of Q Q's remaining mass after 20 days = $w_0$ = initial mass. No decay occurred. Setting up the decay equation: $$w_0 = w_0 \left(\frac{1}{2}\right)^{n_Q} \implies \left(\frac{1}{2}\right)^{n_Q} = 1 \implies n_Q = 0$$ $$n_Q = \frac{t}{t_{1/2}(Q)} = \frac{20}{t_{1/2}(Q)} = 0$$ This is only possible if $t_{1/2}(Q) = \boxed{\infty}$ (infinite) — Q is a stable, non-radioactive isotope. Step 4 - Verify Other Options - **(A) zero**: Instantaneous decay → $w_Q = 0$. Contradicts the 1:4 ratio. Incorrect. - **(B) 5 days**: $n_Q = 4$, $w_Q = w_0/16$. Ratio $w_P/w_Q = (w_0/4)/(w_0/16) = 4:1 \neq 1:4$. Incorrect. - **(C) 20 days**: $n_Q = 1$, $w_Q = w_0/2$. Ratio $= (w_0/4)/(w_0/2) = 1:2 \neq 1:4$. Incorrect. - **(D) infinite**: $w_Q = w_0$ → ratio $= (w_0/4)/w_0 = 1:4$ ✓ **Correct.** $$\boxed{\text{Answer: D — infinite}}$$