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When ₃Li⁷ nuclei are bombarded by protons, and the resultant nuclei are ₄Be⁸, the emitted particles will be
Options
(a) neutrons
(b) alpha particles
(c) beta particles
(d) gamma photons
Correct Answer:
gamma photons
Explanation:
₃Li⁷ + ₁H¹ → ₂Be⁴ + zX ᴬ
Z for the unknown X nucleus = 3 + 1 – 4 = 0
A for the unknown X nucleus = 7 + 1 – 8 = 0
Hence particle emitted has zero Z and zero A
It is a gamma photon.
Related Questions: - Which of the following quantities has units but not dimensions
- Two lenses of power 15 D and -3 D are placed in contact.
- The phase difference between two points seperated by 0.8 m in a wave of frequency
- Young’s modulus of perfectly rigid body material is
- The potential energy of particle in a force field is U = A/r² – b/r, where A and B
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which of the following quantities has units but not dimensions
- Two lenses of power 15 D and -3 D are placed in contact.
- The phase difference between two points seperated by 0.8 m in a wave of frequency
- Young’s modulus of perfectly rigid body material is
- The potential energy of particle in a force field is U = A/r² – b/r, where A and B
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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