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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.
Related Questions: - Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen
- An ideal gas expands in such a manner that its pressure and volume can be related
- The dimensions of universal gas constant are
- A body of mass m is accelerated uniformly from rest to a speed v in a time T
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen
- An ideal gas expands in such a manner that its pressure and volume can be related
- The dimensions of universal gas constant are
- A body of mass m is accelerated uniformly from rest to a speed v in a time T
- When one of the slits of Young’s experiment is covered with a transport sheet
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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