
Successive emission of an $\alpha - $particle and two $\beta - $particles by an atom of a radioactive element results in the formation of its:
(A) Isobar
(B) Isomer
(C) Isotone
(D) Isotope
Answer
232.8k+ views
Hint: In alpha decay, release of a helium nucleus takes place that means the nuclei will observe a decrease in the total mass by 4 while the atomic number will be reduced by 2. However, beta decay is the decay of a neutron to a proton which releases an electron. Due to this, the mass of the atom remains the same but the atomic number is increased by 1.
Complete Step by Step Solution:
In order to become stable, the decay of a radioactive element takes place. Usually, a single decay is not enough for a radioactive nucleus to reach a stable state due to which it undergoes a series of decays until a stable nucleus is formed. There are three types of decays which are described as follows:
Alpha decay: It is a mode of radioactive decay in which a nucleus emits a helium nucleus i.e., two protons and two neutrons.
Beta decay: It is the mode of radioactive decay in which a nucleus emits a beta particle and a release of electrons takes place due to which the nucleus formed after decay has an atomic number higher by the value of 1.
For the given question, let us consider a radioactive element $X$ with atomic mass $A$ and atomic number $Z$. Thus, alpha and beta decay can be represented as follows:
$X_{Z}^{A}\xrightarrow{\alpha }X_{Z-2}^{A-4}\xrightarrow{\beta }X_{Z-1}^{A-4}\xrightarrow{\beta }X_{Z}^{A-4}$
Hence, on successive emissions, the atomic number remains the same but the change is observed in atomic mass. We know that atoms with different atomic mass but same atomic number are referred as isotopes. Therefore, formation isotope takes place after successive emissions.
So, option (D) is the correct answer.
Note: It is important to note that isobars are the atoms with the same atomic mass but different atomic number whereas isotones are the atoms which have the same number of neutrons while number of electrons and protons differ.
Complete Step by Step Solution:
In order to become stable, the decay of a radioactive element takes place. Usually, a single decay is not enough for a radioactive nucleus to reach a stable state due to which it undergoes a series of decays until a stable nucleus is formed. There are three types of decays which are described as follows:
Alpha decay: It is a mode of radioactive decay in which a nucleus emits a helium nucleus i.e., two protons and two neutrons.
Beta decay: It is the mode of radioactive decay in which a nucleus emits a beta particle and a release of electrons takes place due to which the nucleus formed after decay has an atomic number higher by the value of 1.
For the given question, let us consider a radioactive element $X$ with atomic mass $A$ and atomic number $Z$. Thus, alpha and beta decay can be represented as follows:
$X_{Z}^{A}\xrightarrow{\alpha }X_{Z-2}^{A-4}\xrightarrow{\beta }X_{Z-1}^{A-4}\xrightarrow{\beta }X_{Z}^{A-4}$
Hence, on successive emissions, the atomic number remains the same but the change is observed in atomic mass. We know that atoms with different atomic mass but same atomic number are referred as isotopes. Therefore, formation isotope takes place after successive emissions.
So, option (D) is the correct answer.
Note: It is important to note that isobars are the atoms with the same atomic mass but different atomic number whereas isotones are the atoms which have the same number of neutrons while number of electrons and protons differ.
Recently Updated Pages
JEE Main 2026 Session 2 Registration Open, Exam Dates, Syllabus & Eligibility

JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

Trending doubts
JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding Average and RMS Value in Electrical Circuits

Understanding Collisions: Types and Examples for Students

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding Atomic Structure for Beginners

AssertionIn electrolytic refining of metal impure metal class 12 chemistry JEE_Main

Other Pages
JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions Hindi Medium (2025-26)

CBSE Class 12 Chemistry Set 1 56/2/1 2025: Question Paper, Answers & Analysis

CBSE Class 12 Chemistry Question Paper Set 3 2025 with Answers

Inductive Effect and Its Role in Acidic Strength

Degree of Dissociation: Meaning, Formula, Calculation & Uses

