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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: - If two slits in Young’s experiment are 0.4 mm apart and fringe width on a screen
- A transistor is working in common emitter mode. Its amplification factor is 80
- A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If two slits in Young’s experiment are 0.4 mm apart and fringe width on a screen
- A transistor is working in common emitter mode. Its amplification factor is 80
- A circular disc of radius 0.2 meter is placed in a uniform magnetic field of induction
- An electric dipole of dipole moment p is aligned parallel to a uniform electric field E
- Two vessels separately contain two ideal gases A and B at the same temperature,
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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