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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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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A spring balance is attched to the ceiling of lift.A man hangs his bag on the string
- A planet moving along an elliptical orbit is closest to the sun at a distance of r₁
- The dimension of 1/2ε₀E², where ε₀ is permittivity of free space and E is electric field
- Four point masses, each of value m, are placed at the corners of a square ABCD
- A 50Hz AC signal is applied in a circuit of inductance of (1/π)H and resistance 2100Ω
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Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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