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GR8677 #17
Problem
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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},
$

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
$

Deuteron Decay (rare):

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

(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.

Alternate Solutions
 casseverhart132019-09-30 02:48:03 This is utterly surprising! pressure washing OrlandoReply to this comment
casseverhart13
2019-09-30 02:48:03
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 danielsw986672019-10-21 05:38:21 With the given stages of radioactivity, the answer is A. LED bike lights
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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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 :)
 segfault2009-08-22 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.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$