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) \({ }_{0} \mathrm{n}^{1}\) (c) \({ }_{1} \mathrm{H}^{1}\) (d) \({ }_{1} \mathrm{H}^{2}\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Reaction

We are asked to find the projectile used in a nuclear reaction where \( \mathrm{N}^{14} \) is transformed into \( \mathrm{O}^{17} \) and a proton is released. This indicates a nuclear reaction of the form \( \mathrm{X} + \mathrm{N}^{14} \rightarrow \mathrm{O}^{17} + \mathrm{H}^1 \). We need to determine \( \mathrm{X} \).

Step 2: Use Conservation Laws

In nuclear reactions, mass number and atomic number are conserved. For the mass number: \( A_X + 14 = 17 + 1 \). For the atomic number: \( Z_X + 7 = 8 + 1 \). Here, \( A_X \) and \( Z_X \) are the mass number and atomic number of \( \mathrm{X} \), respectively.

Step 3: Solve for Mass Number

From the mass number equation, \( A_X + 14 = 17 + 1 \). Simplify it to \( A_X + 14 = 18 \). Therefore, \( A_X = 18 - 14 = 4 \). The projectile has a mass number of 4.

Step 4: Solve for Atomic Number

From the atomic number equation, \( Z_X + 7 = 8 + 1 \). Simplify it to \( Z_X + 7 = 9 \). Therefore, \( Z_X = 9 - 7 = 2 \). The projectile has an atomic number of 2.

Step 5: Identify the Projectile

With \( A_X = 4 \) and \( Z_X = 2 \), the projectile is \( { }_{2} \mathrm{He}^{4} \). This matches the option (a).

Key Concepts

Projectile Identification

In nuclear reactions, identifying the projectile is crucial as it is the particle that initiates the reaction by colliding with a target nucleus. In the given exercise, \(\mathrm{N}^{14}\) is bombarded, leading to the production of \(\mathrm{O}^{17}\) and the release of a proton. The projectile, therefore, is the missing particle that, when combined with the target, accounts for all the particles in the reaction. This process involves reversing the reaction equation to find the unidentified projectile, labeled \(\mathrm{X}\).

• The initial step is to write the nuclear equation: \(\mathrm{X} + \mathrm{N}^{14} \rightarrow \mathrm{O}^{17} + \mathrm{H}^1\). The goal is to solve for \(\mathrm{X}\).
• We then apply conservation laws, using mass and atomic numbers to deduce properties of \(\mathrm{X}\).
• In this specific problem, calculations showed that \(\mathrm{X}\) corresponds to \(\mathrm{He}\), fitting the given choices with its specific mass and atomic numbers.

Through this approach, identifying the projectile relies on understanding the reaction's context and using logical deductions based on conservation laws.

Mass Number Conservation

Mass number conservation is a fundamental principle in nuclear reactions stating that the total mass number before the reaction equals the total mass number after the reaction. In simpler terms, mass number is conserved. This principle helps in balancing nuclear reactions and verifying unknown particles.
In the reaction: \(\mathrm{X} + \mathrm{N}^{14} \rightarrow \mathrm{O}^{17} + \mathrm{H}^1\), the sum of the mass numbers on both sides must be equal. Initially, this gives us:

• Before the reaction: \(A_{X} + 14\) (mass number of the projectile plus the nitrogen target).
• After the reaction: \(17 + 1\) (mass number of oxygen and the proton).

By calculating, we find that \(A_{X} = 4\).
To students, this demonstrates how each particle's mass contributes to the overall equation, needing coherence to suit the laws of physics.

Atomic Number Conservation

Similar to mass number conservation, atomic number conservation is key in nuclear physics. This principle ensures that the total atomic number remains unchanged during a nuclear reaction, offering a way to solve for unknowns in reactions.
Using atomic number conservation for \(\mathrm{X} + \mathrm{N}^{14} \rightarrow \mathrm{O}^{17} + \mathrm{H}^1\), the equation works as follows:

• Initially, the atomic number equation is \(Z_{X} + 7\) (7 being nitrogen's atomic number).
• After the reaction, it becomes \(8 + 1\) (oxygen plus proton).

Through calculation, \(Z_{X} = 2\).
This solution, where \(Z_{X}\) equates to 2, assists in verifying that the projectile is helium, whose atomic number matches this finding. Atomic number conservation is vital for ensuring all elements in a nuclear reaction adhere to its fundamental rules, simplifying discovery and verification of unknown particles in reactions.