
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
161.1k+ 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 2021 July 25 Shift 1 Question Paper with Answer Key

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

Young's Double Slit Experiment Step by Step Derivation

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

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

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

If a wire of resistance R is stretched to double of class 12 physics JEE_Main
