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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
Related Questions: - Which of the following statements is correct?
- The decay constant of a radio isotope is λ. If A₁ and A₂ are its activities at time
- The masses of two radioactive substances are same and their half lives
- A particle has initial velocity (2i⃗+3j⃗) and acceleration (0.3i⃗+0.2j⃗). The magnitude
- The near point and far point of a person are 40 cm and 250 cm, respectively.
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which of the following statements is correct?
- The decay constant of a radio isotope is λ. If A₁ and A₂ are its activities at time
- The masses of two radioactive substances are same and their half lives
- A particle has initial velocity (2i⃗+3j⃗) and acceleration (0.3i⃗+0.2j⃗). The magnitude
- The near point and far point of a person are 40 cm and 250 cm, respectively.
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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