| ⇦ | 
| ⇨ |
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: - A U²³⁵ reactor generates power at a rate of P producing 2×10⁸ fission per second.
- Maximum velocity of the photoelectrons emitted by a metal surface is 1.2×10⁶ ms⁻¹.
- When a sound wave of wavelength λ is propagating in a medium,
- In an interference experiment, third bright fringe is obtained at a point
- Colours appear on a thin soap film and soap bubbles due to the phenomenon of
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A U²³⁵ reactor generates power at a rate of P producing 2×10⁸ fission per second.
- Maximum velocity of the photoelectrons emitted by a metal surface is 1.2×10⁶ ms⁻¹.
- When a sound wave of wavelength λ is propagating in a medium,
- In an interference experiment, third bright fringe is obtained at a point
- Colours appear on a thin soap film and soap bubbles due to the phenomenon of
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