During the transformation of _aX^b to _cY^d, the number of β-particles emitted are — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Conservation of Mass Number General transformation: $\ce{^b_aX -> ^d_cY} + N_\alpha\,\ce{^4_2He} + N_\beta\,\ce{^0_{-1}e}$ Mass number conservation ($\alpha$ reduces by 4, $\beta$ reduces by 0): $$b = d + 4N_\alpha \implies N_\alpha = \frac{b - d}{4}$$ Step 2 - Conservation of Atomic Number Atomic number conservation ($\alpha$ reduces by 2, $\beta^-$ increases by 1): $$a = c + 2N_\alpha - N_\beta$$ $$N_\beta = c - a + 2N_\alpha = c - a + 2 \cdot \frac{b - d}{4}$$ $$\boxed{N_\beta = (c - a) + \frac{b - d}{2}}$$ Step 3 - Elimination of Wrong Options - (A) $\frac{b-d}{4}$: this is $N_\alpha$, not $N_\beta$ - (B) $(a-c) + \frac{b-d}{2}$: signs reversed — violates atomic number conservation - (C) $(c-a) + \frac{b-d}{2}$: correct derivation ✓ - (D) $(b-d) + 2(c-a)$: incorrectly uses $b-d$ instead of $\frac{b-d}{4}$ for alpha count