
An atomic nucleus \[{}_{90}T{h^{232}}\] emits several a and b radiations and finally reduces to \[{}_{82}P{b^{208}}\]. It must have emitted
A. 4a and 2b
B. 6a and 4b
C. 8a and 24b
D. 4a and 16b
Answer
163.8k+ views
Hint: In this question, we need to find the particles emitted by a and b. For this first, we need to compare the mass numbers of two elements, and then we need to compare the atomic numbers of the same two elements.
Complete step by step solution:Given,
\[{}_{90}T{h^{232}}\] where 232 is the mass number and 90 is the atomic number.
\[{}_{82}P{b^{208}}\] where 208 is the mass number and 82 is the atomic number.
Let a and b be the number of emitted during the reaction
\[{}_{90}T{h^{232}} \to {}_{82}P{b^{208}} + a{}_2H{e^4} + b{}_{ - 1}{e^0}\]
Comparing the mass numbers,
\[232 = 208 + 4a + (bx0)\]
\[4a = 232 - 208\]
\[4a = 24\]
\[a = \frac{{24}}{4}\]
\[a = 6\]
Comparing the atomic numbers
\[90 = 82 + (2xa) + - 1b\]
\[90 = 82 + 2a - 1b\]
\[2a - b = 90 - 82\]
\[2a - b = 8\]
By substituting the value of a=4 in the above equation we get,
\[2x6 - b = 8\]
\[b = 12 - 8\]
\[b = 4\]
So,
The number of a particles emitted=6
The number of b particles emitted =4
Therefore the correct answer is Option B.
Additional Information: \[{}_{90}T{h^{232}}\] belongs to III group. It emits an \[\alpha \]-particle. Because elements from 89 to 103 are positioned in group III. Thorium (chemical symbol Th) is a naturally arising radioactive metal found at trace levels in rocks, soil, plants, water, and animals. Thorium is solid under normal circumstances. There are man-made and natural forms of thorium, all of which are radioactive. Generally, naturally occurring thorium exists like Th-228, Th-230, or Th-232. The nucleus of lead-208 is composed of 82 protons and 126 neutrons and, with a very nice approximation, can be seen as spherical. Lead has four stable isotopes such as Pb-204, Pb-206, Pb-207, Pb-208. Pb-208 is the end of the thorium series from Th-232.
Note: Students should know the atomic and mass number of thorium and lead. If they have any confusion about atomic and mass numbers, they will not get correct and accurate answers. So, students should know the atomic number, mass number, and name of the elements.
Complete step by step solution:Given,
\[{}_{90}T{h^{232}}\] where 232 is the mass number and 90 is the atomic number.
\[{}_{82}P{b^{208}}\] where 208 is the mass number and 82 is the atomic number.
Let a and b be the number of emitted during the reaction
\[{}_{90}T{h^{232}} \to {}_{82}P{b^{208}} + a{}_2H{e^4} + b{}_{ - 1}{e^0}\]
Comparing the mass numbers,
\[232 = 208 + 4a + (bx0)\]
\[4a = 232 - 208\]
\[4a = 24\]
\[a = \frac{{24}}{4}\]
\[a = 6\]
Comparing the atomic numbers
\[90 = 82 + (2xa) + - 1b\]
\[90 = 82 + 2a - 1b\]
\[2a - b = 90 - 82\]
\[2a - b = 8\]
By substituting the value of a=4 in the above equation we get,
\[2x6 - b = 8\]
\[b = 12 - 8\]
\[b = 4\]
So,
The number of a particles emitted=6
The number of b particles emitted =4
Therefore the correct answer is Option B.
Additional Information: \[{}_{90}T{h^{232}}\] belongs to III group. It emits an \[\alpha \]-particle. Because elements from 89 to 103 are positioned in group III. Thorium (chemical symbol Th) is a naturally arising radioactive metal found at trace levels in rocks, soil, plants, water, and animals. Thorium is solid under normal circumstances. There are man-made and natural forms of thorium, all of which are radioactive. Generally, naturally occurring thorium exists like Th-228, Th-230, or Th-232. The nucleus of lead-208 is composed of 82 protons and 126 neutrons and, with a very nice approximation, can be seen as spherical. Lead has four stable isotopes such as Pb-204, Pb-206, Pb-207, Pb-208. Pb-208 is the end of the thorium series from Th-232.
Note: Students should know the atomic and mass number of thorium and lead. If they have any confusion about atomic and mass numbers, they will not get correct and accurate answers. So, students should know the atomic number, mass number, and name of the elements.
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 Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

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

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

NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

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
