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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: - A conductor of length 5 cm is moved parallel to itself with a speed of 2 m/s,
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A conductor of length 5 cm is moved parallel to itself with a speed of 2 m/s,
- A thin semicircular conducting ring (PQR) of radius r is falling with its plane
- A super conductor exhibits perfect
- The radioactivity of a certain material drops to 1/16 of the initial value in 2 hours.
- A resistance R draws power P when connected to an AC source. If an inductance
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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