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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
- A point charge q is situated at a distance r on axis from one end of a thin
- The equivalent resistance of two resistors connected in series is 6 Ω
- An object moving with a speed of 6.25 m/s, is deaccelerated at a rate given by
- A black body radiates 20 W at temperature 227⁰C. It temperature of the black body
- A closely wound solenoid of 2000 turns and area of cross-section 1.5 x 10⁻⁴ m² carries
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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