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

positron, proton, neutron, neutron.

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Advanced 2011 · Shift 1 · Q11 with Solution

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Correct answer: A

Identify the known artificial disintegration reactions of aluminium by $\alpha$ -particles

Aluminium is $^{27}_{13}\text{Al}$ and an $\alpha$ -particle is $^{4}_{2}\text{He}$ .

So the initial combined nucleus is:

^{27}_{13}\text{Al} + ^{4}_{2}\text{He} \Rightarrow A=31,\ Z=15

Thus, after bombardment, the products must conserve:

Mass number = $31$
Atomic number = $15$

First mode of disintegration

A well-known reaction is:

^{27}_{13}\text{Al} + ^{4}_{2}\text{He} \rightarrow ^{30}_{15}\text{P} + ^{1}_{0}n

Check conservation:

Mass: $27+4=30+1=31$
Atomic number: $13+2=15+0=15$

So in one path, the emitted particle is a neutron.

Hence one of $X, Y, Z$ must be neutron.

Second mode of disintegration

Another known reaction is:

^{27}_{13}\text{Al} + ^{4}_{2}\text{He} \rightarrow ^{30}_{14}\text{Si} + ^{1}_{1}p

Check conservation:

Mass: $27+4=30+1=31$
Atomic number: $13+2=14+1=15$

So another emitted particle is a proton.

Thus among $X, Y, Z$ , another product is proton.

Formation of positron

The phosphorus formed above, $^{30}_{15}\text{P}$ , is radioactive and decays by positron emission:

^{30}_{15}\text{P} \rightarrow ^{30}_{14}\text{Si} + ^{0}_{+1}e

So the third product is a positron.

Match with options

Thus the three products $X, Y, Z$ are:

\text{proton, neutron, positron}

This matches Option A.

Final answer

The correct option is:

\boxed{\text{A: proton, neutron, positron}}

More from Chemical Kinetics and Nuclear Chemistry

View written solutionFree

Correct answer: A

Identify the known artificial disintegration reactions of aluminium by $\alpha$ -particles

Aluminium is $^{27}_{13}\text{Al}$ and an $\alpha$ -particle is $^{4}_{2}\text{He}$ .

So the initial combined nucleus is:

^{27}_{13}\text{Al} + ^{4}_{2}\text{He} \Rightarrow A=31,\ Z=15

Thus, after bombardment, the products must conserve:

Mass number = $31$
Atomic number = $15$

First mode of disintegration

A well-known reaction is:

^{27}_{13}\text{Al} + ^{4}_{2}\text{He} \rightarrow ^{30}_{15}\text{P} + ^{1}_{0}n

Check conservation:

Mass: $27+4=30+1=31$
Atomic number: $13+2=15+0=15$

So in one path, the emitted particle is a neutron.

Hence one of $X, Y, Z$ must be neutron.

Second mode of disintegration

Another known reaction is:

^{27}_{13}\text{Al} + ^{4}_{2}\text{He} \rightarrow ^{30}_{14}\text{Si} + ^{1}_{1}p

Check conservation:

Mass: $27+4=30+1=31$
Atomic number: $13+2=14+1=15$

So another emitted particle is a proton.

Thus among $X, Y, Z$ , another product is proton.

Formation of positron

The phosphorus formed above, $^{30}_{15}\text{P}$ , is radioactive and decays by positron emission:

^{30}_{15}\text{P} \rightarrow ^{30}_{14}\text{Si} + ^{0}_{+1}e

So the third product is a positron.

Match with options

Thus the three products $X, Y, Z$ are:

\text{proton, neutron, positron}

This matches Option A.

Final answer

The correct option is:

\boxed{\text{A: proton, neutron, positron}}

Chemical Kinetics and Nuclear Chemistry question

Chemical Kinetics and Nuclear Chemistry question

More from Chemical Kinetics and Nuclear Chemistry

Chemical Kinetics and Nuclear Chemistry question

Chemical Kinetics and Nuclear Chemistry question

More from Chemical Kinetics and Nuclear Chemistry