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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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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Angle of contact of a liquid with a solid depends on
- If the image formed by a convex mirror of focal length 30 cm is a quarter of the size
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- A particle of mass m is projected with velocity v making an angle of 45° with
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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