| ⇦ | 
| ⇨ |
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: - Power dissipated in an LCR series circuit connected to an a.c source of emf ? is
- If temperature of gas increases from 27°C to 927°C, the kinetic energy will be
- Angle of minimum deviation for a prism of reractive index 1.5 equal to the angle
- The electromagnetic theory of light failed to explain
- A stone thrown into still water, creates a circular wave pattern moving radially
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Power dissipated in an LCR series circuit connected to an a.c source of emf ? is
- If temperature of gas increases from 27°C to 927°C, the kinetic energy will be
- Angle of minimum deviation for a prism of reractive index 1.5 equal to the angle
- The electromagnetic theory of light failed to explain
- A stone thrown into still water, creates a circular wave pattern moving radially
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