| ⇦ |
| ⇨ |
When ₉₀Th²²⁸ transforms to ₈₃Bi²¹², the number of emitted α and β particles are respectively
Options
(a) 8 α,7 β
(b) 4 α,7 β
(c) 4 α,4 β
(d) 4 α,1 β
Correct Answer:
4 α,1 β
Explanation:
α-particle = ₂He⁴, β-particle = ₋₁β and Nucleus = zXᴬ
Change in A occurs only due to α-emission.
Change in A = 228 – 212 = 16
This change is due to 4 α.
Again change in Z = 90 – 83 = 7
Change in Z due to 4α = 8
.·. Change in Z due to β = 8 – 7 = 1
This is due to one β.
Hence particles emitted = 4α, 1β.
Related Questions: - A metal wire of circular cross-section has a resistance R₁. The wire is now stretched
- If two bodies are projected at 30⁰ and 60⁰ respectively with the same velocity, then
- Length of a conductor is 50 cm radius of cross section 0.1 cm and resistivity
- The physical quantities not having same dimensions are
- The number of photo electrons emitted for light of a frequency v
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A metal wire of circular cross-section has a resistance R₁. The wire is now stretched
- If two bodies are projected at 30⁰ and 60⁰ respectively with the same velocity, then
- Length of a conductor is 50 cm radius of cross section 0.1 cm and resistivity
- The physical quantities not having same dimensions are
- The number of photo electrons emitted for light of a frequency v
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Leave a Reply