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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: - The electric potential at a point in free space due to a charge Q coulomb is Q x 10¹¹ v
- A charge Q is uniformly distributed over a large plastic plate. The electric field
- Maximum velocity of the photoelectrons emitted by a metal surface is 1.2×10⁶ ms⁻¹.
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The electric potential at a point in free space due to a charge Q coulomb is Q x 10¹¹ v
- A charge Q is uniformly distributed over a large plastic plate. The electric field
- Maximum velocity of the photoelectrons emitted by a metal surface is 1.2×10⁶ ms⁻¹.
- The angular speed of earth, so that the object on equator may appear weightless
- An α-particle and a proton travel with the same velocity in a magnetic field
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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