Question

When \({ }_{13} \mathrm{Al}^{27}\) is bombarded with \(\alpha\) particle, a radioactive isotope of phosphorus \({ }_{15} \mathrm{P}^{30}\) is formed. Which particle is emitted along with \({ }_{15} \mathrm{P}^{30-}\) ? (a) deuteron (b) proton (c) electron (d) neutron

Short answer

The particle emitted is a neutron.

Step-by-step solution

Identify the Starting Materials

We start with \( _{13} \text{Al}^{27} \) and an \( \alpha \) particle. An \( \alpha \) particle is essentially a helium nucleus, \( _{2}\text{He}^{4} \).

Identify the Product

The problem states that a radioactive isotope of phosphorus, specifically \( _{15} \text{P}^{30} \), is formed from the bombardment of the aluminum nucleus by the \( \alpha \) particle.

Write the Nuclear Reaction Equation

We can write the nuclear reaction as: \[ _{13} \text{Al}^{27} + _{2} \text{He}^{4} \rightarrow _{15} \text{P}^{30} + X \]Here \( X \) is the unknown particle that is emitted.

Apply the Conservation of Mass Number

The mass numbers on both sides of the reaction equation must balance. Left side: \( 27 + 4 = 31 \)Right side: \( 30 + A \)Solving for \( A \), we find:\[ 31 = 30 + A \Rightarrow A = 1 \]

Apply the Conservation of Atomic Number

The atomic numbers on both sides of the reaction equation must also balance.Left side: \( 13 + 2 = 15 \)Right side: \( 15 + Z \)Solving for \( Z \), we find:\[ 15 = 15 + Z \Rightarrow Z = 0 \]

Identify the Particle X

The particle \( X \) must have a mass number \( A = 1 \) and an atomic number \( Z = 0 \). This corresponds to a neutron (\( _{0} \text{n}^{1} \)).

Confirm the Emitted Particle

The emitted particle (\( X \)) is a neutron, consistent with \( A = 1 \) and \( Z = 0 \). Thus, the correct answer is \( \text{(d) neutron} \).

Key concepts

Alpha Particle Bombardment

Alpha particle bombardment is a process where a target nucleus is struck by alpha particles, typically causing nuclear transformations. An alpha particle is essentially the nucleus of a helium atom, composed of two protons and two neutrons. It is denoted as \( _{2} \text{He}^{4} \), indicating it has an atomic number of 2 and a mass number of 4.

When an alpha particle bombards a nucleus, various nuclear reactions can take place, often resulting in the formation of a new element and the emission of another particle. In our exercise, \( _{13} \text{Al}^{27} \), an aluminum nucleus, is bombarded by an alpha particle, resulting in the creation of \( _{15} \text{P}^{30} \), a phosphorus isotope, among other things.

This process helps in studying nuclear structure and properties. It provides insights into nuclear reactions that occur naturally in stars and during radioactive decay.

Some common results of alpha particle bombardment include:

• Transformation into different elements.
• Emission of new particles such as neutrons or protons.
• Changes in the energy state of the nucleus.

Conservation of Mass Number

The conservation of mass number is a fundamental principle in nuclear reactions. It states that the total mass number before and after a nuclear reaction must remain the same. The mass number (A) of an atom is the sum of protons and neutrons in its nucleus.

In the given exercise, the starting materials are \( _{13} \text{Al}^{27} \) and \( _{2} \text{He}^{4} \), giving a total mass number before the reaction of \( 27 + 4 = 31 \). After the bombardment, the mass numbers must sum up to the same total. Here's how the balance is maintained:

We identify that phosphorus \( _{15} \text{P}^{30} \) is formed, and to balance the reaction:

• Total mass number before: 31
• Total mass number after: 30 + A

Here, solving for \( A \), we find it to be 1. This confirms that a particle with mass number 1 is also emitted, ensuring the mass number conservation.

This principle is crucial because it reflects the immutable nature of protons and neutrons in reactions, crucially standing alongside the conservation of atomic numbers.

Conservation of Atomic Number

The conservation of atomic number is another key principle, ensuring that the total number of protons before and after a nuclear reaction remain unchanged. The atomic number (Z) of an atom represents the number of protons in its nucleus.

For our exercise, we look at the aluminum nucleus \( _{13} \text{Al}^{27} \) and an alpha particle \( _{2} \text{He}^{4} \) on the left side of the equation. This gives:

• Total atomic number before reaction: 13 + 2 = 15
• Target atomic number after reaction: 15, due to phosphorus \( _{15} \text{P}^{30} \)

To ensure the balance, any additional particles emitted must bring the total to 15.

By solving the equation \( 15 = 15 + Z \), we find that \( Z \) must be 0. This identifies the particle emitted as one having an atomic number of 0, a neutron \( _{0} \text{n}^{1} \). Neutrons do not alter the atomic number since they do not carry charge.

Understanding this conservation helps in predicting and explaining the products of nuclear reactions, reflecting how particles rearrange but keep the total proton count constant.