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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
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Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The angular velocity of a wheel increases from 100 rps to 300 rps in 10 seconds.
- A spring balance is attached to the ceiling of a lift.A man hangs his bag on the spring
- The periodic waves of intensities I₁ and I₂ pass through a region at the same time
- Eight drops of a liquid of density ρ and each of radius a are falling through air
- A 36 Ω galvanometer is shunted by resistance of 4 Ω. The percentage of the total current
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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