Problem 142 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the following nuclear reactions \({ }_{92} \mathrm{M}^{238} \longrightarrow \mathrm{y}^{\mathrm{N}}^{\mathrm{x}}+2{ }_{2} \mathrm{He}^{4}\) \({ }_{\mathrm{y}} \mathrm{N}^{\mathrm{x}} \longrightarrow{\mathrm{B}} \mathrm{L}^{\mathrm{A}}+2 \beta^{+}\) The number of neutrons in element \(\mathrm{L}\) is (a) 146 (b) 144 (c) 142 (d) 140

Step-by-Step Solution

Answer

The number of neutrons in element L is 144, so the answer is (b) 144.

Step 1: Understand the initial reaction

Start with the equation for the nuclear reaction: \( { }_{92} \text{M}^{238} \rightarrow \mathrm{y}^{\mathrm{N}}^{\mathrm{x}}+2{ }_{2} \text{He}^{4}. \) We know that when a nucleus emits an alpha particle \([}^4_2{\text{He}}],\) the atomic number decreases by 2 and the mass number decreases by 4. Since there are two alpha particles, the atomic number of \(\text{M}\) must be reduced by 4, and the mass number by 8.

Step 2: Calculate Y and X for N

Calculate the new atomic and mass numbers for element \(\mathrm{N}\): initially the element \({ }_{92} \text{M}^{238} \) emits two \(^4_2{\text{He}}\), which means: \(\mathrm{y} = 92 - 2 \times 2 = 88\) and \(\mathrm{x} = 238 - 2 \times 4 = 230.\) Thus, \(\mathrm{N}\) is represented as \({ }_{88} N^{230}.\)

Step 3: Analyze the second nuclear reaction

The second reaction is \({ }_{88} N^{230} \rightarrow { }_{\mathrm{b}} \mathrm{L}^{\mathrm{A}} + 2 \beta^{+},\) where each \(\beta^{+}\) emission decreases the atomic number by 1 (without changing the mass number). Since two \(\beta^{+}\) particles are emitted, the atomic number decreases by 2, meaning \(\mathrm{b} = 88 - 2 = 86.\) The mass number is unchanged, so \(\mathrm{A} = 230.\)

Step 4: Calculate the number of neutrons in L

The number of neutrons in \(\text{L}\) can be calculated using the formula: Neutrons = Mass number - Atomic number. Therefore, \(\text{neutrons in L} = \text{A} - \text{b} = 230 - 86 = 144.\)

Key Concepts

Alpha Decay

Alpha decay is a type of radioactive decay where a nucleus emits an alpha particle.
This alpha particle is essentially a helium nucleus, consisting of two protons and two neutrons, denoted as \(^4_2\text{He}\).
Alpha decay helps unstable nuclei achieve a more stable state by reducing their atomic and mass numbers.When a nucleus undergoes alpha decay:

• The atomic number decreases by 2 because it loses 2 protons.
• The mass number decreases by 4 due to the loss of 2 protons and 2 neutrons.

For example, if a nucleus starts as \(_{92}\text{M}^{238}\), emitting two alpha particles would decrease the atomic number by 4 \((2\times2)\) and the mass number by 8 \((2\times4)\).This process transforms the isotope into a completely different element, noted for its altered atomic number and mass.

Beta Decay

Beta decay is another form of radioactive decay but different from alpha decay.
In this process, a beta particle, either an electron \((\beta^-)\) or a positron \((\beta^+)\), is emitted from the nucleus.
Our example uses positron emission, resulting in a decrease of the atomic number.During positron emission:

• The atomic number decreases by 1 because a proton in the nucleus is converted into a neutron.
• The mass number remains unchanged as the total number of nucleons (protons + neutrons) is conserved.

In the original exercise, \({ }_{88} \text{N}^{230}\) emits two positrons, decreasing the atomic number of the resultant \(\text{L}\) to 86, while the mass number stays at 230. This demonstrates how beta decay contributes to the transformation of elements without altering the mass number.

Neutron Calculation

Calculating the number of neutrons in an atom is crucial for understanding its stability and characteristics.
Neutrons are uncharged particles found in the nucleus, accompanying protons.
To find the number of neutrons, use the formula:\[\text{Neutrons} = \text{Mass Number} - \text{Atomic Number}\]In the given exercise for the element \(\text{L}\), we know:

• Mass number (A) = 230
• Atomic number (b) = 86

Hence, the number of neutrons:\[230 - 86 = 144\]This number is important because the presence of more or fewer neutrons can affect the stability of the nucleus and the type of decay it might undergo.

Atomic and Mass Number

The atomic and mass numbers are fundamental attributes of any atom, revealing its type and structure.
The atomic number (Z) is the number of protons in the nucleus, defining the element.
The mass number (A) is the total number of protons and neutrons in the atom.Changing either of these numbers transforms the identity of the element:

• Atomic Number: Dictates the chemical element. For example, Atomic number 92 corresponds to Uranium.
• Mass Number: Indicates isotopes which have the same number of protons but different numbers of neutrons. Thus, \(^1\text{H}\) is hydrogen, \(^2\text{H}\) is deuterium.

Understanding these values is key when analyzing nuclear reactions like in the exercise where alpha and beta decay alter these numbers to form new elements like \(\text{L}\).

Radioactive Decay Processes

Radioactive decay processes are natural changes in the nucleus of an atom that lead to the emission of particles or electromagnetic radiation.
These processes help unstable isotopes reach a stable form.
The most common types include alpha, beta, and gamma decay. Why do these occur?

• Atoms have excess energy or mass, making them unstable.
• Decay processes release this energy for stabilization.

In the given example, the original atom undergoes alpha decay followed by beta decay:

• Alpha Decay: Reduces both atomic and mass numbers, creating a new element.
• Beta Decay: Alters the atomic number by converting neutrons to protons or vice versa, also transforming the element identity.

These decays illustrate the diverse paths atoms can take to achieve stability.