GR | # Login | Register
  GR8677 #17
GREPhysics.NET Official Solution    Alternate Solutions
Verbatim question for GR8677 #17
Advanced Topics}Radioactivity

X^A_Z\rightarrow ? \rightarrow ? \rightarrow Y^{A-4}_{Z-1} by natural decay
Recall the following decay processes, and recall that the subscripts and superscripts have to balance on both sides of the arrow.

Beta (\beta^-) Decay:

where \bar{\nu} is the symbol of an anti-neutrino. (The \beta^+ decay is accompanied by a neutrino, \nu, instead of an anti-neutrino.)

Alpha Decay:

Gamma Decay (no change in A or Z):

Deuteron Decay (rare):

(A) X^A_Z\rightarrow X^A_{Z+1} + \beta^0_{-1} \rightarrow X'^{A-4}_{Z-1} +He^4_2 \rightarrow Y^{A-4}_{Z-1}. This works out.

(B) \beta^- decay always occurs with an antineutrino.

(C) \gamma emission doesn't change the atomic number.

(D) Although the numbers work out here, deuteron decay is rare, thus not natural.

(E) The subscripts and superscripts don't add up.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
2019-09-30 02:48:03
This is utterly surprising! pressure washing OrlandoAlternate Solution - Unverified
2019-09-30 02:48:03
This is utterly surprising! pressure washing Orlando
2019-10-21 05:38:21
With the given stages of radioactivity, the answer is A. LED bike lights
Alternate Solution - Unverified
2019-08-10 03:09:30
I would say that muon is a partical from Lepton group so are electron, tauon, neutrino. So the answer must be (A). lol player searchNEC
2019-08-08 07:20:47
Well, I am so excited that I have found this your post because I have been searching for some information about it almost three hours. You helped me a lot indeed and reading this your article I have found many new and useful information about this subject. air ductNEC
2009-08-12 21:37:41
I believe ETS made a typo in the problem statement. The subscript on Y should be Z+1 to be consistent with the rest of the problem.
That is because if Z (the # of protons) decreases by 1, it implies a proton turning into a neutron (\beta^+ decay). To conserve charge you must emit a positron and to conserve lepton number (from the positron) you must also emit an electron neutrino.

In short, p -> n + e^+ +\nu_e

Correct me if I'm wrong :)
2009-08-22 22:06:46
After the *first* stage in the process the subscript should be Z+1, but after alpha emission the subscript becomes Z+1-2=Z-1. This is a two step process.

Post A Comment!
Click here to register.
This comment is best classified as a: (mouseover)
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .
type this... to get...
$\langle my \rangle$
$\left( abacadabra \right)_{me}$
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...