When ₃Li⁷ nuclei are bombarded by protons, and the resultant nuclei are ₄Be⁸,

⇦

⇨

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:

1. A tuning fork gives 4 beats with 50 cm length of a sonometer
2. Minority carriers in a p-type semiconductor are
3. The pair of quantities having same dimensions is
4. The material suitable for making electromagnets should have
5. An inductor 20 mH, a capacitor 50 μF and a resistor 40Ω are connected in series

Topics: Atoms and Nuclei (136)
Subject: Physics (2479)

---

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

---

Share this page with your friends