Problem 133 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following Column-I (Reactions) (a) \({ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+\ldots \ldots\) (b) \({ }_{6} \mathrm{C}^{12}+\ldots \ldots \rightarrow{ }_{5} \mathrm{~B}^{10}+{ }_{2} \mathrm{He}^{4}\) (c) \({ }_{7} \mathrm{~N}^{14}+\ldots \ldots \rightarrow{ }_{8} \mathrm{O}^{17}+{ }_{1} \mathrm{H}^{1}\) (d) \({ }_{20} \mathrm{Ca}^{40}+\ldots \ldots \rightarrow{ }_{19} \mathrm{~K}^{37}+{ }_{2} \mathrm{He}^{4}\) Column-II (Particles) (p) \({ }_{2} \mathrm{He}^{4}\) (q) \({ }_{0} \mathrm{n}^{1}\) (r) \({ }_{1} \mathrm{D}^{2}\) (s) \({ }_{1} \mathrm{H}^{1}\)

Step-by-Step Solution

Answer

(a) matches (q), (b) matches (r), (c) matches (p), (d) matches (s).

Step 1: Analyze Reaction (a)

Examine the nuclear reaction: \( {}_{4} \mathrm{Be}^{9}+{}_{2} \mathrm{He}^{4} \rightarrow{}_{6} \mathrm{C}^{12}+\ldots \ldots \). Compare the atomic mass and atomic numbers on both sides of the reaction. Here, the left side has a total of 13 protons and 4 + 9 = 13 in mass. Thus, the missing particle must be a neutron to conserve both mass and charge: \( {}_{0} \mathrm{n}^{1} \).

Step 2: Analyze Reaction (b)

Examine the nuclear equation: \( {}_{6} \mathrm{C}^{12}+\ldots \ldots \rightarrow{}_{5} \mathrm{~B}^{10}+{}_{2} \mathrm{He}^{4} \). Carbon has 12 as mass and boron plus helium sum to 14, so the missing particle should be a deuteron \( {}_{1} \mathrm{D}^{2} \).

Step 3: Analyze Reaction (c)

Evaluate the equation: \( {}_{7} \mathrm{~N}^{14}+\ldots \ldots \rightarrow{}_{8} \mathrm{O}^{17}+{}_{1} \mathrm{H}^{1} \). It suggests an addition of 3 mass units and 1 proton difference between nitrogen and oxygen. The missing particle must be \( {}_{2} \mathrm{He}^{4} \).

Step 4: Analyze Reaction (d)

Review the equation: \( {}_{20} \mathrm{Ca}^{40}+\ldots \ldots \rightarrow{}_{19} \mathrm{~K}^{37}+{}_{2} \mathrm{He}^{4} \). The calcium loses 3 in mass and 1 proton, so it requires a particle that is \( {}_{1} \mathrm{H}^{1} \) or proton to maintain conservation rules.

Key Concepts

Conservation of Mass and Charge

In nuclear reactions, the conservation of mass and charge is a fundamental principle. This principle asserts that the total number of protons and the total mass must remain constant from beginning to end of the reaction. Consider it a balancing act between the reactants and products involved in the nuclear change.

• Mass Conservation: The sum of the mass numbers (protons plus neutrons) on the reactants' side must equal the sum on the products' side.
• Charge Conservation: The sum of the atomic numbers (protons) on each side of the equation must also be equal.

This means that when you're writing nuclear equations, the left and right sides should align perfectly in both mass and charge. For instance, if a heavier nucleus emits a neutron to become a lighter nucleus, we must account for the balance in mass and add a neutron with zero charge to the products. Using this approach helps to track and match each particle in the equation accurately.

Proton and Neutron Identification

Understanding how to identify protons and neutrons is crucial when analyzing nuclear reactions. Each element is defined by its atomic number, which equals the number of protons. Neutrons, on the other hand, are found by subtracting the atomic number from the mass number. Both play significant roles in maintaining nuclear stability.

Let's break it down:

• Protons: These are positively charged particles found in the nucleus of every atom. The number of protons defines an element's identity. For example, Carbon has 6 protons.
• Neutrons: These particles have no charge and are also located in the nucleus. They act as a glue to hold the nucleus together. Their number can vary in atoms of the same element, leading to different isotopes.

During nuclear reactions, you must be able to count these particles to ensure they are conserved. This involves determining and verifying their numbers from both sides of a nuclear equation. This approach helps in deducing any missing particles in a reaction, like identifying a neutron in the product of a reaction.

Nuclear Equations Analysis

Analyzing nuclear equations involves more than just rearranging numbers; it requires understanding the intricacies of how particles transform. When examining an equation, consider the changes in atomic numbers and mass numbers, since these shifts indicate the types of particles involved.

Here’s how to analyze nuclear equations:

• Identify Given Particles: Start by noting all given nuclei and particles on each side of the equation.
• Check Conservation: Ensure that mass and charge are conserved by calculating totals for both sides.
• Determine Missing Particles: If something is unaccounted for, use the conservation rules to deduce what type of particle must be present.

For example, in the reaction \( {}_{4} \mathrm{Be}^{9} + {}_{2} \mathrm{He}^{4} \rightarrow {}_{6} \mathrm{C}^{12} + \ldots \ldots \), you'll discover that a neutron \( {}_{0} \mathrm{n}^{1} \) completes both mass and charge balance. This analytical skill allows you to untangle what seems initially complex and assists in learning how nuclear reactions are formed and resolved.