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By the succesive disintegration of ₉₂U²³⁸, the final product obtained is ₈₂Pb²⁰⁶, then how many number of α and β-particles are emitted?
Options
(a) 6 and 8
(b) 8 and 6
(c) 12 and 6
(d) 8 and 12
Correct Answer:
8 and 6
Explanation:
The number of α-particles, n₁ = Change in mass number / 4
n₁ = [(238 – 206) / 4] = 32 / 4 = 8
Now, number of β-particles, n₂ = 82 – (92 – 2 n₁)
= 82 – (92 – 2 × 8) = 82 – 76 = 6
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A piece of iron is heated in a flame. It first comes dull red then becomes reddish
- A potentiometer wire of length 10 m and resistance 10 Ω per metre is connected in series
- A proton is moving in a uniform magnetic field B in a circular path of radius a
- A body projected vertically from the earth reaches a height equal to earth’s radius
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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