Problem 33 - A Complete Resource Book in Chemistry for JEE Main

Question

The projectile used to bombard, \(\mathrm{N}^{14}\) to get \({ }_{8} \mathrm{O}^{17}\) and a proton is (a) \({ }_{2} \mathrm{He}^{4}\) (b) \(_{0} n^{1}\) (c) \({ }_{1} \mathrm{H}^{1}\) (d) \({ }_{1} \mathrm{H}^{2}\)

Step-by-Step Solution

Answer

The projectile is _{2} He^{4}. (Option a)

Step 1: Understand the Nuclear Reaction

In this problem, we have a nuclear reaction where \(N^{14}\) is bombarded by a particle to produce \(O^{17}\) and a proton. This can be represented by a nuclear equation: \[\mathrm{N}^{14} + X \rightarrow \mathrm{O}^{17} + \mathrm{p}\] where \(X\) is the unknown projectile.

Step 2: Determine the Properties of the Unknown Projectile

To find the unknown projectile \(X\), we need to use the conservation of mass number and atomic number or charge in nuclear reactions. The total mass numbers and atomic numbers before and after the reaction must be equal.

Step 3: Apply Conservation of Mass Number

The mass number of \(\mathrm{N}^{14}\) is 14, and the mass number of \(\mathrm{O}^{17}\) is 17. A proton (\(\mathrm{p}\)) has a mass number of 1. Therefore, the equation for conservation of mass number is: \[14 + A = 17 + 1\] where \(A\) is the mass number of the projectile.

Step 4: Solve for Mass Number

Rearrange the equation to find \(A\): \[A = 17 + 1 - 14 = 4\] So, the mass number of the projectile \(X\) is 4.

Step 5: Apply Conservation of Atomic Number

For conservation of atomic number, nitrogen (\(N\)) has an atomic number of 7 and oxygen (\(O\)) has an atomic number of 8. A proton has an atomic number of 1. Therefore, the equation is: \[7 + Z = 8 + 1\] where \(Z\) is the atomic number of the projectile.

Step 6: Solve for Atomic Number

Rearrange the equation to find \(Z\): \[Z = 8 + 1 - 7 = 2\] So, the atomic number of the projectile \(X\) is 2.

Step 7: Identify the Projectile

The projectile has an atomic number of 2 and a mass number of 4. This corresponds to \(\mathrm{He}^4\), the helium nucleus, also known as an alpha particle. Thus, \(X\) is \(\,_{2}\mathrm{He}^{4}\).

Key Concepts

Conservation of Mass Number

Nuclear reactions must follow certain fundamental rules, and one of these is the conservation of mass number. The mass number represents the sum of protons and neutrons in an atomic nucleus. In a nuclear reaction, it's a rule that the total mass number of the reactants must equal the total mass number of the products.

For instance, consider the nuclear equation: \[^{14}\text{N} + X \rightarrow \text{O}^{17} + \text{p} \]. In this scenario:

• \(^{14}\text{N}\) has a mass number of 14.
• \(\text{O}^{17}\) has a mass number of 17.
• A proton (\(\text{p}\)) has a mass number of 1.

To find the mass number of the unknown particle \(X\), we balance the equation: \[14 + A = 17 + 1 \]. Solving this gives \(A = 4\), indicating that the mass number of \(X\) is 4. This preservation of mass number ensures the physical integrity of matter in nuclear reactions.

Conservation of Atomic Number

Another basic principle in nuclear reactions is the conservation of atomic number. The atomic number is crucial because it determines the identity of the element; essentially, it is the number of protons in the nucleus.

Using the same nuclear reaction: \(^{14}\text{N} + X \rightarrow \text{O}^{17} + \text{p} \):

• Nitrogen (N) has an atomic number of 7.
• Oxygen (O) has an atomic number of 8.
• A proton itself represents an atomic number of 1.

For conservation, we set up the equation: \[7 + Z = 8 + 1\]. Solving for \(Z\), we find \(Z = 2\).

This consistency in atomic number underscores the reliability and predictability of nuclear reactions, as each reaction reflects an exact swap in the nucleus without affecting the overall identity of cosmic elements.

Alpha Particle

An alpha particle may sound complex, but it’s simply a helium nucleus! Alpha particles are a type of ionizing radiation ejected from the cores of certain nuclear materials during radioactive decay. They consist of:

• 2 protons
• 2 neutrons

This gives them a mass number of 4 and an atomic number of 2. When identifying the unnamed particle \(X\) in the nuclear reaction, we found:

• Mass number = 4
• Atomic number = 2

These characteristics perfectly match an alpha particle, making \(X\) the alpha particle \(\,_{2}\text{He}^{4}\). Alpha particles are notable for their strong interaction with matter, which allows them to cause significant ionization along their path but means they travel only short distances due to their mass and double positive charge.

Projectile Identification

In nuclear reactions, identifying the projectile is like piecing together a puzzle to understand the whole picture. The projectile is the initial particle that strikes another to trigger the reaction.

In our case, the task is to determine \(X\) in the equation \(^{14}\text{N} + X \rightarrow \text{O}^{17} + \text{p} \). By applying both conservation laws:

• The mass number conservation: led to an unknown particle with \(A = 4\).
• The atomic number conservation: required \(Z = 2\).

These characteristics pointed to an alpha particle \(\,_{2}\text{He}^{4}\). Bombardment with an alpha particle is a common technique in nuclear physics experiments to study nuclear reactions or create new elements and isotopes. Understanding the role and identity of projectiles helps scientists predict the outcomes of nuclear encounters and design further explorations into atomic structures.