Question

Bombardment of aluminium be \(\alpha\) -particle leads to its artificial disintegration in two way (i) and (ii) as shown. Products \(\mathrm{X}, \mathrm{Y}\) and \(Z\), respectively, are \({ }_{13}^{27} \mathrm{Al} \longrightarrow{ }_{14}^{30} \mathrm{Si}+\mathrm{X}\) \({ }_{13}^{27} \mathrm{Al} \stackrel{\text { (ii) }}{\longrightarrow}{ }_{15}^{30} \mathrm{P}+\mathrm{Y}\) \({ }_{15}^{30} \mathrm{P} \longrightarrow{ }_{14}^{30} \mathrm{Si}+\mathrm{Z}\) (a) proton, neutron, positron (b) neutron, positron, proton (c) proton, positron, neutron (d) positron, proton, neutron

Short answer

There is an error in the given answer choices, as none match the actual disintegration products: electron (beta-minus particle), alpha particle, and positron, respectively.

Step-by-step solution

Conservation of Atomic Number

Firstly, we must account for the conservation of the atomic number in the nuclear reactions (i) and (ii). The atomic number must be conserved before and after the reaction. For reaction (i), the atomic number on the left is 13 (for Al), and on the right, it is 14 (for Si). Whatever product X is, it should have an atomic number of -1 to maintain the conservation, since 13 + (-1) = 14. For reaction (ii), the atomic number on the left is also 13 (for Al), and on the right, it is 15 (for P). Thus, product Y should have an atomic number of -2 to conserve the atomic number, since 13 + (-2) = 15.

Conservation of Mass Number

Next, the conservation of mass number should also be considered. The mass number before and after the reaction must be equal. For reaction (i), the mass number on the left is 27, and on the right, it is 30. This means product X must have a mass number of -3 (27 - 3 = 30). For reaction (ii), the mass number is the same on both sides, so product Y must have a mass number of 0. Lastly, in reaction (iii), mass number is conserved because 30 is found on both sides of the equation, implying product Z has a mass number of 0.

Identification of Disintegration Products

From steps 1 and 2, we concluded that product X has an atomic number of -1 and a mass number of -3; thus, X must be a beta particle (specifically a beta-minus particle or an electron). Product Y, with an atomic number of -2, must be a particle that contains two protons which is an alpha particle. Finally, product Z, having no change in mass number and atomic number from P to Si, must be a positron which is consistent with the beta-plus decay process.

Conclusion of the Nuclear Reaction Products

Given the conservation laws and the identifications from step 3, the products of the artificial disintegration can be concluded as follows:\(\mathrm{X} \) is a beta particle or an electron (not mentioned in the options but is implied by atomic number -1), \(\mathrm{Y} \) is an alpha particle, and \(\mathrm{Z} \) is a positron. Substituting the usual names for these particles into the conservation equations, we find that the products are: (a) proton, neutron, positron (incorrect, as X is not a proton); (b) neutron, positron, proton (incorrect, as X is not a neutron); (c) proton, positron, neutron (incorrect, as Y is not a positron and X is not a proton); (d) positron, proton, neutron (incorrect as per Step 3's results). Since the given options do not accurately represent the disintegration products, there seems to be an error in the options.

Key concepts

Conservation of Atomic Number

Understanding the conservation of atomic number is crucial when studying nuclear reactions. It states that the total number of protons in a nuclear reaction must remain constant before and after the reaction occurs. This is a bedrock principle in chemistry and physics, as it helps us predict the outcome of radioactive decay processes like alpha and beta decay.

When an alpha particle strikes the nucleus of an aluminum atom, as shown in exercise (i), it transforms into silicon, altering the atomic number. The product X must account for the difference to conserve the atomic number. Likewise, in exercise (ii), the atomic number increases by 2 when phosphorus is formed from aluminum. The role of conservation of atomic number is straightforward, yet it provides an essential check that helps us identify the particles involved in nuclear reactions.

Conservation of Mass Number

The conservation of mass number in nuclear physics requires that the sum of protons and neutrons, which constitute the mass number, remains unchanged through the course of a nuclear reaction. This allows us to understand changes occurring within the nucleus, where the total mass is more than just the sum of its parts due to nuclear binding energy.

For the reactions given, conservation of mass number dictates what types of particles can be produced. In reaction (i) and (iii), we notice that the mass number increases and remains constant, respectively, compelling us to reconcile these changes with appropriate particles that carry or have no mass number. This principle aids us in deducing the identity of disintegration products by ensuring that the mass remains accounted for.

Alpha and Beta Decay

Alpha and beta decay are two types of radioactive decay processes that play a pivotal role in nuclear chemistry. Alpha decay involves the release of an alpha particle, which is essentially a helium nucleus with 2 protons and 2 neutrons. This type of decay decreases the mass number by 4 and the atomic number by 2.

Beta decay, on the other hand, involves the transformation of a neutron into a proton, accompanied by the emission of an electron (beta-minus decay) or a proton into a neutron with the emission of a positron (beta-plus decay). Beta-minus decay increases the atomic number by 1 without changing the mass number, whereas beta-plus decay decreases the atomic number by 1, also without altering the mass number. Understanding these decay processes is instrumental in identifying the particles emitted during nuclear reactions, as shown in the given exercise.