Help | Contact Us
NukeWorker Menu

Gollnick problem... can't solve

Started by Zorro1, Nov 30, 2015, 03:05

Previous topic - Next topic

0 Members and 1 Guest are viewing this topic.

Zorro1

I know the solution to this is probably simple but on problem set 9 of ch. 2 in Gollnick's Basic rad pro tech. it asks: Calculate the average energy of the negatron and neutrino emitted in the beta decay of 90Y..... the answers are (Beta = 0.758 MeV, neutrino = 1.52MeV)    how did they come to this?

Rennhack

Good question, thank you for asking.

Chapter 2, Page 44:
Quote
In beta decay, we often call this energy the endpoint energy or Emaximum. On the average, betas carry about 1/3 of the endpoint energy.

2.278 * 1/3 = 0.758


Chapter 2, Page 46:
Quote
Remember that the betas emitted have a range of energies up to the value of the decay energy, Emaximum. On the average, they carry 1/3 Emax and the neutrinos 2/3 Emax.

2.278 * 2/3 = 1.52


NukeWorker ™ is a registered trademark of NukeWorker.com ™, LLC © 1996-2025 All rights reserved.
All material on this Web Site, including text, photographs, graphics, code and/or software, are protected by international copyright/trademark laws and treaties. Unauthorized use is not permitted. You may not modify, copy, reproduce, republish, upload, post, transmit or distribute, in any manner, the material on this web site or any portion of it. Doing so will result in severe civil and criminal penalties, and will be prosecuted to the maximum extent possible under the law.
Privacy Statement | Terms of Use | Code of Conduct | Spam Policy | Advertising Info | Contact Us | Forum Rules | Password Problem?