| ⇦ | 
| ⇨ |
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: - If capillary experiment is performed in vaccum, then a liquid under the condition
- A pure semiconductor behaves slightly as a conductor at
- CO₂ laser uses
- Electrons in a certain energy level n=n₁, can emit 3 spectral lines. When they
- Two rigid bodies A and B rotate with rotational kinetic energies Eᴀ and Eʙ
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If capillary experiment is performed in vaccum, then a liquid under the condition
- A pure semiconductor behaves slightly as a conductor at
- CO₂ laser uses
- Electrons in a certain energy level n=n₁, can emit 3 spectral lines. When they
- Two rigid bodies A and B rotate with rotational kinetic energies Eᴀ and Eʙ
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
Leave a Reply