| ⇦ | 
| ⇨ |
₉₂U²³⁵ undergoes successive disintegrations with the end product of ₈₂P²⁰³. The number of α and β-particles emitted are
Options
(a) α=8,β=6
(b) α=3,β=3
(c) α=6,β=4
(d) α=6,β=0
Correct Answer:
α=8,β=6
Explanation:
₉₂U²³⁵ → end product ₈₂P²°³ α and β emitted.
ΔA = 235 – 203 = 32
Therefore, 8 alpha particles are emitted. The charge should be 92 – 16 = 76.
But as the final charge is 82, six β⁻ particles had been emitted to make up the final atomic number Z = 82.
.·. 8 alpha particles and six β⁻ have been emitted.
Related Questions: - A car moving with a speed of 50 kmh⁻¹ can be stopped by brakes after at least 6 m
- Pascal-second has the dimensions of
- A short circuited coil is placed in a magnetic field varying with time. Electrical
- When photons of energy hν are incident on the surface of photosensitive
- The input signal given to a CE amplifier having a voltage gain of 150 is
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A car moving with a speed of 50 kmh⁻¹ can be stopped by brakes after at least 6 m
- Pascal-second has the dimensions of
- A short circuited coil is placed in a magnetic field varying with time. Electrical
- When photons of energy hν are incident on the surface of photosensitive
- The input signal given to a CE amplifier having a voltage gain of 150 is
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