Bombardment of aluminium be α-particle leads to its artificial disintegration in two way (i) and (ii — Nuclear Chemistry and Radioactivity Chemistry Question
Question
Bombardment of aluminium be α-particle leads to its artificial disintegration in two way (i) and (ii) as shown. Products X, Y and Z, respectively, are _13Al^27 --(i)--> _14Si^30 + X, _13Al^27 --(ii)--> _15P^30 + Y, _15P^30 -> _14Si^30 + Z
💡 Solution & Explanation
Step 1 - Write Both Bombardment Reactions $$\ce{^{27}_{13}Al + ^{4}_{2}He ->[\text{(i)}] ^{30}_{14}Si + X}$$ $$\ce{^{27}_{13}Al + ^{4}_{2}He ->[\text{(ii)}] ^{30}_{15}P + Y}$$ $$\ce{^{30}_{15}P -> ^{30}_{14}Si + Z}$$ Step 2 - Find X from Pathway (i) Conserve charge: $13 + 2 = 14 + z_X \implies z_X = 1$ Conserve mass: $27 + 4 = 30 + a_X \implies a_X = 1$ $z=1, a=1$ → **X = proton** ($\ce{^{1}_{1}H}$) Step 3 - Find Y from Pathway (ii) Conserve charge: $13 + 2 = 15 + z_Y \implies z_Y = 0$ Conserve mass: $27 + 4 = 30 + a_Y \implies a_Y = 1$ $z=0, a=1$ → **Y = neutron** ($\ce{^{1}_{0}n}$) Step 4 - Find Z from P-30 Decay $\ce{^{30}_{15}P}$ is proton-rich (positron emitter): Conserve charge: $15 = 14 + z_Z \implies z_Z = 1$ but as positron: $z_Z = +1$ (positron $e^+$) Conserve mass: $30 = 30 + a_Z \implies a_Z = 0$ $z=+1, a=0$ → **Z = positron** ($\ce{e^+}$, or $\beta^+$) Step 5 - Evaluate Options - **(A) proton, neutron, positron**: X=proton ✓, Y=neutron ✓, Z=positron ✓. **Correct.** - **(B) neutron, positron, proton**: Incorrect assignment. - **(C) proton, positron, neutron**: Incorrect assignment. - **(D) positron, proton, neutron**: Incorrect assignment. $$\boxed{\text{Answer: A — X = proton, Y = neutron, Z = positron}}$$