⇦ | ⇨ |
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: - The power obtained in a reactor using U²³⁵ disintegration is 1000 kW
- A car moving with a velocity of 36 km/hr crosses a siren of frequency 500 Hz.
- We get Balmer series when electron comes from
- A projectile is thrown in the upward direction making an angle of 60°with the horizontal
- The electric potential at the surface of an atomic nucleus (z=50) of radius
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The power obtained in a reactor using U²³⁵ disintegration is 1000 kW
- A car moving with a velocity of 36 km/hr crosses a siren of frequency 500 Hz.
- We get Balmer series when electron comes from
- A projectile is thrown in the upward direction making an angle of 60°with the horizontal
- The electric potential at the surface of an atomic nucleus (z=50) of radius
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