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₉₂U²³⁴ has 92 protons and 238 nucleons. It decays by emitting an alpha particle and becomes
Options
(a) ₉₂U²³⁴
(b) ₉₀Th²³⁴
(c) ₉₂U²³⁵
(d) ₉₃Np²³⁷
Correct Answer:
₉₀Th²³⁴
Explanation:
Emission of alpha particle, decreases the mass number by 4 and charge number by 2.
.·. Decrease in mass number = 238 – 4 = 234
Decrease in charge number = 92 – 2 = 90 ₉₀Th²³⁴ is emitted.
Related Questions: - By the succesive disintegration of ₉₂U²³⁸, the final product obtained is ₈₂Pb²⁰⁶, then
- A black body at 227⁰C radiates heat at the rate of 7cals/cm²s. At a temperature
- All components of the electromagnetic spectrum in vacuum have the same
- The rest mass of a body is m1. It moves with a velocity of 0.6c, then its relativistic
- When a mass m is attached to a spring, it normally extends by 0.2 m.
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- By the succesive disintegration of ₉₂U²³⁸, the final product obtained is ₈₂Pb²⁰⁶, then
- A black body at 227⁰C radiates heat at the rate of 7cals/cm²s. At a temperature
- All components of the electromagnetic spectrum in vacuum have the same
- The rest mass of a body is m1. It moves with a velocity of 0.6c, then its relativistic
- When a mass m is attached to a spring, it normally extends by 0.2 m.
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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