A radioactive element \[{}_{90}{X^{238}}\] decay into \[{}_{83}{Y^{222}}\]. The number of \[\beta \] particles emitted are
A. 4
B. 6
C. 2
D. 1
Answer
292.2k+ views
Hint:This problem can be solved by using the concept of radioactive decay. There are three types of radioactive decay. They are alpha decay, beta decay and gamma decay. By applying alpha decay and beta decay concepts, we can find the resultant product.
Complete step by step solution:
Change in mass number (A) = 238 – 222 = 16
Number of \[\alpha \]-particles emitted = \[\dfrac{{238 - 222}}{4} = \dfrac{{16}}{4} = 4\]
After the emission of \[\alpha \]-particles, the change in atomic number will be \[90 - (4 \times 2) = 82\].
Since the atomic number of $Y$ is 83 and this is possible only when there is one beta decay.
Therefore, the number of \[\beta \] particles emitted = 1
Hence, the correct option is option D.
Additional Information: An atom's nuclei exhibit radioactivity as a result of nuclear instability. Henry Becquerel was the first person to discover radioactivity in 1986. The decay of atomic nuclei with the emission of helium nuclei or electrons and high-energy photons is called radioactivity.
The three types of radioactivity are given below.
1) Alpha decay: The decay of atomic nuclei with the emission of helium nuclei is called alpha decay. The alpha particle has two protons and two neutrons. The equation for alpha decay is given below.
\[{}_{Z}{X^{A}}\longrightarrow{}_{Z-2}{Y^{A-4}}+{}_{2}{He^{4}}\]
2) Beta decay: The decay of atomic nuclei with the emission of an electron or positron is called beta decay. A beta particle is generally referred to as an electron but it can also be a positron also. The equation for beta decay is given below.
\[{}_{Z}{X^{A}}\longrightarrow{}_{Z+1}{Y^{A}}+{}_{-1}{e^{0}}\]
3) Gamma decay: The decay of atomic nuclei with the emission of photons is called gamma decay. During the gamma decay process, neither A nor Z changes. The equation for gamma decay is given below.
\[({}_{Z}{X^{A}})^*\longrightarrow{}_{Z}{X^{A}}+\gamma\]
Note: In beta decay there are two types. Negative beta decay releases a negatively charged beta particle called an electron. Positive beta decay releases a positively charged beta particle called a positron.
Complete step by step solution:
Change in mass number (A) = 238 – 222 = 16
Number of \[\alpha \]-particles emitted = \[\dfrac{{238 - 222}}{4} = \dfrac{{16}}{4} = 4\]
After the emission of \[\alpha \]-particles, the change in atomic number will be \[90 - (4 \times 2) = 82\].
Since the atomic number of $Y$ is 83 and this is possible only when there is one beta decay.
Therefore, the number of \[\beta \] particles emitted = 1
Hence, the correct option is option D.
Additional Information: An atom's nuclei exhibit radioactivity as a result of nuclear instability. Henry Becquerel was the first person to discover radioactivity in 1986. The decay of atomic nuclei with the emission of helium nuclei or electrons and high-energy photons is called radioactivity.
The three types of radioactivity are given below.
1) Alpha decay: The decay of atomic nuclei with the emission of helium nuclei is called alpha decay. The alpha particle has two protons and two neutrons. The equation for alpha decay is given below.
\[{}_{Z}{X^{A}}\longrightarrow{}_{Z-2}{Y^{A-4}}+{}_{2}{He^{4}}\]
2) Beta decay: The decay of atomic nuclei with the emission of an electron or positron is called beta decay. A beta particle is generally referred to as an electron but it can also be a positron also. The equation for beta decay is given below.
\[{}_{Z}{X^{A}}\longrightarrow{}_{Z+1}{Y^{A}}+{}_{-1}{e^{0}}\]
3) Gamma decay: The decay of atomic nuclei with the emission of photons is called gamma decay. During the gamma decay process, neither A nor Z changes. The equation for gamma decay is given below.
\[({}_{Z}{X^{A}})^*\longrightarrow{}_{Z}{X^{A}}+\gamma\]
Note: In beta decay there are two types. Negative beta decay releases a negatively charged beta particle called an electron. Positive beta decay releases a positively charged beta particle called a positron.
Recently Updated Pages
JEE Main 2025-26 Electrostatics Mock Test: Free Practice Online

Electrostatics Mock Test 2025: Practice Questions & Answers

JEE Main 2026-27 Kinematics Mock Test – Free Practice Online

JEE Main Mock Test 2025-26: Chapter-Wise Practice Papers

Wheatstone Bridge – Principle, Formula, Diagram & Applications

Kinematics Mock Test 2025: Free Practice Questions & Answers

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

Understanding Uniform Acceleration in Physics

What Are Current and Potential Difference in Electricity?

