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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 circular disc X of radius R is made from an iron plate of thickness t and another disc
- Sodium has body centred packing. Distance between two nearest atoms is 3.7 Å.
- The wave function (in SI unit) for a light wave is given as Ψ(x,t)= 10³ sin π
- A car moves from X to Y with a uniform speed v(u) and returns to Y with a uniform speed
- The range of voltmeter is 10V and its internal resistance is 50Ω.
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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