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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.
Related Questions: - A uranium nucleus ₉₂ U ²³⁸ emits an α-particle and a β-particle in succession.
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Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A uranium nucleus ₉₂ U ²³⁸ emits an α-particle and a β-particle in succession.
- The diameter of the eye ball of a normal eye is about 2.5 cm. The power of the eye
- Two circuits have a mutual inductance of 0.1 H. What average e.m.f. is induced
- Suppose for some reasons, radius of earth were to shrink by 1% of present value,
- Application of Bernoulli’s theorem can be seen in
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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