| ⇦ |
| ⇨ |
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: - A tuning fork A produces 4 beats per second with another tuning fork B of frequency
- If the length of a closed organ pipe is 1.5 m and velocity of sound is 330 m/s,
- You are marooned on a frictionless horizontal plane and cannot exert any horizontal
- A milli ammeter of range 10 mA has a coil of resistance 1 ohm
- The Kα X-ray of molybdenum has a wavelength of 71×10⁻¹² m. If the energy of a molybdenum
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A tuning fork A produces 4 beats per second with another tuning fork B of frequency
- If the length of a closed organ pipe is 1.5 m and velocity of sound is 330 m/s,
- You are marooned on a frictionless horizontal plane and cannot exert any horizontal
- A milli ammeter of range 10 mA has a coil of resistance 1 ohm
- The Kα X-ray of molybdenum has a wavelength of 71×10⁻¹² m. If the energy of a molybdenum
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Leave a Reply