
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
161.4k+ 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 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main Mock Test Series Class 12 Chemistry for FREE

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Classification of Drugs

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Solutions Class 12 Notes: CBSE Chemistry Chapter 1

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

Electrochemistry Class 12 Notes: CBSE Chemistry Chapter 2
