Problem 62 - A Complete Resource Book in Chemistry for JEE Main

Question

If a \(\mathrm{X}^{\mathrm{b}}\) species emits firstly a positron, then \(2 \alpha\) and \(2 \beta\) particles and in last \(1 \alpha\) particle is also emitted and finally converts to \(\mathrm{Y}^{\mathrm{d}}\) species, so the correct relation is (a) \(\mathrm{c}=\mathrm{a}-5, \mathrm{~d}=\mathrm{b}-12\) (b) \(\mathrm{c}=\mathrm{a}-5, \mathrm{~d}=\mathrm{b}-10\) (c) \(\mathrm{c}=\mathrm{a}-6, \mathrm{~d}=\mathrm{b}-0\) (d) \(\mathrm{c}=\mathrm{a}-4, \mathrm{~d}=\mathrm{b}-12\)

Step-by-Step Solution

Answer

Option (a) is correct: \( \mathrm{c} = \mathrm{a} - 5, \mathrm{d} = \mathrm{b} - 12 \).

Step 1: Initial Setup

First, understand the transformations involved. The species \( \mathrm{X}^{\mathrm{b}} \) emits particles and transforms into \( \mathrm{Y}^{\mathrm{d}} \). We need to find the relationship between the atomic number (\( \mathrm{c} \)) and mass number (\( \mathrm{d} \)) for \( \mathrm{Y} \), with respect to \( \mathrm{X} \)'s atomic (\( \mathrm{a} \)) and mass number (\( \mathrm{b} \)).

Step 2: First Positron Emission

A positron emission will decrease the atomic number by 1 but leaves the mass number unchanged. Thus, the new atomic number is \( \mathrm{a} - 1 \) and the mass number remains \( \mathrm{b} \).

Step 3: Emitting Two \( \alpha \) Particles

Each \( \alpha \) particle decreases the atomic number by 2 and mass number by 4. Emitting 2 \( \alpha \) particles decreases the atomic number by 4 and mass number by 8, giving an updated atomic number of \( \mathrm{a} - 5 \) and mass number of \( \mathrm{b} - 8 \).

Step 4: Emitting Two \( \beta \) Particles

Each \( \beta \) particle emission increases the atomic number by 1, without changing the mass number. Emitting 2 \( \beta \) particles increases the atomic number by 2, resulting in \( \mathrm{a} - 3 \) while the mass number remains \( \mathrm{b} - 8 \).

Step 5: Final \( \alpha \) Particle Emission

Emitting 1 more \( \alpha \) particle decreases the atomic number by 2 and the mass number by 4, leading to a final atomic number of \( \mathrm{a} - 5 \) and mass number of \( \mathrm{b} - 12 \).

Step 6: Conclusion

The final species \( \mathrm{Y}^{\mathrm{d}} \) has \( \mathrm{c} = \mathrm{a} - 5 \) and \( \mathrm{d} = \mathrm{b} - 12 \). Match these results with the given options to find that option (a) is correct.

Key Concepts

Alpha Decay

In nuclear physics, alpha decay is a process where an unstable atom emits an alpha particle to become a more stable atom. An alpha particle consists of two protons and two neutrons, which is essentially a helium nucleus.
When an atom undergoes alpha decay, several changes occur:

• Atomic number decreases by 2: Since the alpha particle has two protons, the element loses two protons from its nucleus, effectively changing the element to one that is two places earlier in the Periodic Table.
• Mass number decreases by 4: The mass number represents the total number of protons and neutrons. With two protons and two neutrons ejected, the mass number also drops by four.

Understanding alpha decay is essential because it affects both the atomic and mass numbers significantly by reducing them in a straightforward way.

Beta Decay

Beta decay is another common form of radioactive decay. During this process, a neutron in the nucleus of an atom is transformed into a proton, and a beta particle (electron or positron) is emitted.
There are important changes to note:

• Atomic number increases by 1: In the case of negative beta decay (emission of an electron), the neutron-turned-proton increases the atomic number by one, changing the element to the next one on the Periodic Table.
• Mass number remains unchanged: Despite the increase in the atomic number, the overall mass number remains constant because the change involves a neutron turning into a proton without loss of total nucleons.

Beta decay is crucial in nuclear reactions because it increases the atomic number without affecting the mass number.

Positron Emission

Positron emission is a type of beta decay where a proton in the nucleus is converted into a neutron, emitting a positron—the antimatter counterpart of the electron. This process is key in adjusting atomic structures.
Here’s what happens during positron emission:

• Atomic number decreases by 1: Because a proton changes to a neutron, the atomic number decreases by one, effectively moving the element one place lower in the Periodic Table.
• Mass number remains unchanged: Like other forms of beta decay, the total mass number stays the same as the neutron-proton transformation does not impact the overall nucleon count.

Positron emission is vital for understanding processes where an atom needs to reduce its atomic number without changing its mass.

Atomic Number Changes

The atomic number of an element is fundamental because it determines the identity of the element within the Periodic Table. It is defined as the number of protons present in the nucleus of an atom. During various nuclear reactions, atomic numbers can change in several ways:

• Decreases: In alpha decay or positron emission, the atomic number can decrease due to the loss of protons.
• Increases: Beta decay increases the atomic number as neutrons convert into protons.

Understanding how and why atomic number changes occur is crucial for predicting the new elements formed by nuclear reactions.

Mass Number Changes

The mass number represents the total count of nucleons (protons and neutrons) within an atom’s nucleus. It affects the atom's weight and stability. Nuclear reactions can cause variations in the mass number:

• Decreases: Alpha decay reduces the mass number by four due to the ejection of an alpha particle consisting of two protons and two neutrons.
• Unchanged: During beta decay and positron emission, the mass number remains constant as only internal nucleon transformations occur without net loss.

Comprehending mass number changes helps in understanding the isotopes resultant from nuclear reactions and their subsequent behavior.