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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.
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle is released from rest from a tower of height h, i.e,t₁ : t₂ : t₃ is
- The momentum of a photon of energy 1 MeV in kg m/s will be
- Perfectly black body radiates the energy 18 J/s at 300K. Another ordinary body of e=0.8
- A current of 5 A is passing through a metallic wire of cross-sectional area 4×10⁻⁶ m².
- The work done in which of the following processes is equal to the change in internal energy
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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