• anonymous
I have small doubt: In compound nucleus theory, why the decay mode is independent of the formation mode.
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • jamiebookeater
I got my questions answered at in under 10 minutes. Go to now for free help!
  • anonymous
The simplest explanation is that if formed a compound nucleus has a sufficient lifetime to exist as a excited state of the coalition of the projectile and target. Dthis time is long compared to the transit time of the projectile. During this time the energy within the compound nucleus is distributed throughout the nucleus thus loosing any correlation with the initiating reaction. The compound nucleus with its own unique quantum number then decays to allowed final states.

Looking for something else?

Not the answer you are looking for? Search for more explanations.