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

Question

The projectile used to bombard \({ }_{7} \mathrm{~N}^{14}\) to get \({ }_{8} \mathrm{O}^{17}\) and a proton is (a) \({ }_{2} \mathrm{He}^{4}\) (b) \({ }_{0} \mathrm{n}^{1}\) (c) \({ }_{1} \mathrm{H}^{1}\) (d) \({ }_{1} \mathrm{H}^{2}\)

Step-by-Step Solution

Answer

The projectile used is \( {}_{2} \text{He}^{4} \).

Step 1: Understanding the Reaction

The given nuclear reaction involves bombarding a nitrogen-14 nucleus, represented as \({ }_{7} ext{N}^{14}\), with a projectile to form an oxygen-17 nucleus, \({ }_{8} ext{O}^{17}\), and a proton, denoted as \({}_{1} ext{H}^{1}\). We need to identify the projectile used.

Step 2: Setting up the Reaction Equation

The nuclear reaction can be represented as:\[ {}_{7} \text{N}^{14} + X \rightarrow {}_{8} \text{O}^{17} + {}_{1} \text{H}^{1} \]where \(X\) is the unknown projectile.

Step 3: Balancing Nuclear Reactions

To identify \(X\), we need to ensure that the sum of the atomic and mass numbers on both sides of the equation are equal. The atomic numbers: \(7 + Z_X = 8 + 1\), giving us \(Z_X = 2\).The mass numbers: \(14 + A_X = 17 + 1\), giving us \(A_X = 4\).

Step 4: Identifying the Projectile

The projectile \(X\) must have atomic number 2 and mass number 4, which corresponds to the helium nucleus, also known as an alpha particle: \({}_{2} \text{He}^{4}\). The nuclear reaction can now be written as:\[ {}_{7} \text{N}^{14} + {}_{2} \text{He}^{4} \rightarrow {}_{8} \text{O}^{17} + {}_{1} \text{H}^{1} \].

Key Concepts

Projectile Identification

In nuclear reactions, identifying the projectile is crucial for understanding what initiates the transformation of nuclei. In our scenario, we start with a nitrogen-14 nucleus and end up with an oxygen-17 nucleus and a proton. The challenge is to figure out what particle bombarded the nitrogen to cause this transformation.

To determine the identity of the projectile, we represent it as an unknown, 'X', in the reaction equation like this:

• Initial nucleus: \( {}_{7} \text{N}^{14} \)
• Product nucleus: \( {}_{8} \text{O}^{17} \)
• Emitted particle: \( {}_{1} \text{H}^{1} \)
• Projectile: \( X \)

Our task is to find the values of atomic and mass numbers for the unknown 'X' so that the equation balances. This identification helps us pinpoint that the projectile in our exercise is the helium nucleus, \( {}_{2} \text{He}^{4} \), often referred to as an alpha particle.

Atomic Number Balancing

Balancing nuclear reactions involves ensuring that the sum of atomic numbers—those little numbers at the bottom—are equal on both sides of the equation. This rule honors the conservation of electrical charge during a nuclear reaction.

Let's glance at the equation with our nitrogen and unknown projectile 'X':

• Initial reaction side: \( 7 + Z_X \)
• Final reaction side: \( 8 + 1 \) (for oxygen-17 and a proton)
• To balance, set: \( 7 + Z_X = 8 + 1 \)
• Solving gives: \( Z_X = 2 \)

The atomic number of '2' tells us we are dealing with helium as the projectile. This process highlights the essential concept that atomic numbers must be logically reconciled to maintain the integrity of the nuclear reaction.

Mass Number Balancing

In nuclear equations, mass number balancing ensures the total mass is the same before and after a reaction. Think of the mass number as the total count of protons and neutrons in a nucleus.

In our case, the mass numbers for the equation are structured like this:

• Initial reaction side: \( 14 + A_X \) (from nitrogen and unknown 'X')
• Final reaction side: \( 17 + 1 \) (from oxygen-17 and a proton)
• To balance, solve: \( 14 + A_X = 17 + 1 \)
• This gives: \( A_X = 4 \)

This matching implies that our unknown projectile, having a mass number of 4, is indeed a helium nucleus. Such balanced equations demonstrate that both sides of a nuclear reaction should reflect equivalent mass, reinforcing the law of mass conservation in nuclear physics.