C^14 is a β active nucleus. A sample of C^14H_4 gas kept in a closed vessel shows increase in pressu — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Beta-Minus Decay of Carbon-14 $\ce{^{14}_6C}$ undergoes $\beta^-$ decay: $$\ce{^{14}_6C -> ^{14}_7N + ^0_{-1}e + \bar{\nu}_e}$$ The carbon nucleus converts to nitrogen ($Z = 6 \to 7$). Step 2 - Chemical Transformation of CH4 Molecule When the $\ce{^{14}C}$ in methane decays to $\ce{^{14}N}$, the molecule converts: $$\ce{^{14}CH4 -> [^{14}NH4^+] + e^- + \bar{\nu}_e}$$ The unstable $\ce{^{14}NH4^+}$ ion rearranges in the gas phase: $$\ce{^{14}NH4^+ -> ^{14}NH3(g) + H^+} \implies \ce{^{14}NH4+ + e^- -> ^{14}NH3(g) + \frac{1}{2}H2(g)}$$ Overall: $\ce{^{14}CH4(g) -> ^{14}NH3(g) + \frac{1}{2}H2(g)}$ Step 3 - Mole Count and Pressure By Ideal Gas Law at constant $V$ and $T$: $P \propto n$ $$n_{\text{initial}} = 1\text{ mol CH}_4 \implies n_{\text{final}} = 1\text{ mol NH}_3 + 0.5\text{ mol H}_2 = 1.5\text{ mol}$$ Moles increase $1 \to 1.5$ → pressure increases. ✓ Step 4 - Option Analysis - (A) $\ce{^{14}NH3}$ and $\ce{H2}$: correct — $n/p$ increase gives nitrogen, ammonia rearrangement yields H2 ✓ - (B) $\ce{B^{11}H3}$ and $\ce{H2}$: incorrect — boron ($Z=5$) is not produced by C-14 decay - (C) $\ce{C^{14}_2H4}$ and $\ce{H2}$: incorrect — ignores nuclear transmutation to nitrogen - (D) $\ce{C^{12}H3}$, $\ce{N^{14}H2}$, $\ce{H2}$: incorrect — C-12 not produced from C-14 $$\boxed{A}$$