GREPhysics.NET: GR8677 #17

GR8677 #17

Problem

GREPhysics.NET Official Solution

Advanced Topics $\Rightarrow$ }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:

$X^A_Z\rightarrow X'^A_{Z+1} + \beta^0_{-1} +\bar{\nu}, <br />$

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:

$X^A_Z\rightarrow X'^{A-4}_{Z-2}+He^{4}_{2}$

Gamma Decay (no change in A or Z):

$X^A_Z\rightarrow X^A_Z\rightarrow + \gamma <br />$

Deuteron Decay (rare):

$X^A_Z\rightarrow X^{A-2}_{Z-1}+H^2_1<br />$

(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.

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Comments

ernest21
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 search

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joshuaprice153
2019-08-08 07:20:47

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jmason86
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 :)

segfault
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.

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