| ⇦ | 
| ⇨ |
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: - In some region, gravitational field is zero. The gravitational potential
- The potential energy of particle in a force field is U = A/r² – b/r, where A and B
- When a beam of light is used to determine the position of an object,
- A beam of light consisting of red, green and blue colours is incident on a right
- The de-Broglie wavelength of a proton(charge = 1.6 x 10⁻¹⁹ C, Mass = 1.6 x 10⁻²⁷ kg)
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In some region, gravitational field is zero. The gravitational potential
- The potential energy of particle in a force field is U = A/r² – b/r, where A and B
- When a beam of light is used to determine the position of an object,
- A beam of light consisting of red, green and blue colours is incident on a right
- The de-Broglie wavelength of a proton(charge = 1.6 x 10⁻¹⁹ C, Mass = 1.6 x 10⁻²⁷ kg)
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