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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
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An a.c. supply gives 30 volt r.m.s which passes through a 10Ω resistance.
- The conductivity in the intrinsic semiconductor does not depend on
- If in a L-R series circuit the power factor is 1/2 and R=100Ω, then the value of L is,
- The ratio of the radii or gyration of a circular disc to that of a circular ring
- A fringe width of a certain interference pattern is β=0.002 cm. What is the distance
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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