When ₃Li⁷ nuclei are bombarded by protons, and the resultant nuclei are ₄Be⁸, the emitted particles will be (a) alpha particles (b) beta particles (c) gamma photons (d) neutrons

Answer: (c). ₃Li⁷ + ₁H¹ → ₄Be⁸ + X A: 8 = 8 + AX ⇒ AX = 0; Z: 4 = 4 + ZX ⇒ ZX = 0 A particle with A = 0, Z = 0 is a photon. So the answer is (c) gamma photons.

When ₃Li⁷ nuclei are bombarded by protons, and the resultant nuclei… | JEE Main 2006 (AIEEE)

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Nuclear Reactions & Q-value

When ${}_{3}Li^{7}$ nuclei are bombarded by protons, and the resultant nuclei are ${}_{4}Be^{8}$ , the emitted particles will be

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Answer: (c)

${}_{3}Li^{7} + {}_{1}H^{1} \to {}_{4}Be^{8} + X$
$A$ : $8 = 8 + A_X \Rightarrow A_X = 0$ ; $Z$ : $4 = 4 + Z_X \Rightarrow Z_X = 0$
A particle with $A = 0$ , $Z = 0$ is a photon.
So the answer is (c) gamma photons.

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