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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.
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Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which component of electromagnetic spectrum have maximum wavelength?
- SI unit of permittivity is
- A particle of mass m is driven by a machine that delivers a constant power k watts
- The force of repulsion between two electrons at a certain distance is F.
- If λ₁ and λ₂ are the wavelengths of the first members of the Lyman and Paschen
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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