IIT-JEE 2011 · Paper 1 · Chemistry Q3

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ChemistrySingle correct+3 / -1Question 3 of 69

Bombardment of aluminum by $\alpha$ -particle leads to its artificial disintegration in two ways, (i) and (ii) as shown. Products X, Y and Z respectively are,

Scheme: ²⁷₁₃Al → (i) ³⁰₁₄Si + X; ²⁷₁₃Al → (ii) ³⁰₁₅P + Y, and ³⁰₁₅P → ³⁰₁₄Si + Z.

proton, neutron, positron

neutron, positron, proton

proton, positron, neutron

positron, proton, neutron

Answer and worked solution

Official answer: A

Balance A and Z ( $^{27}_{13}$ Al + $^4_2$ He: A = 31, Z = 15).

(i) → $^{30}_{14}$ Si + X: X has A = 1, Z = 1, a proton.
(ii) → $^{30}_{15}$ P + Y: Y has A = 1, Z = 0, a neutron.
$^{30}_{15}$ P → $^{30}_{14}$ Si + Z: Z has A = 0, Z = +1, a positron.

(A)

Topics:Radioactive decay kinetics

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IIT-JEE 2011 · Paper 1, Chemistry question 3