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I have to answer this on my worksheet. I understand nuclear chemistry equations a bit, as well as the process of alpha decay and somewhat about what happens when an atom changes its number of protons and neutrons, I understand the gamma symbol and the m symbol meaning the element is high-energy state but this has me stumped. It's an 8 step question. It reads:

The inhalation of radon-222 and its decay to form other isotopes poses a health hazard. Write balanced nuclear equations for the decay of radon-222 to lead-206 in eight steps. Show step 1 below, and show steps 2-8 are on the next page.

a. Step 1: radon-222 decays by alpha emission.

b. Step 2: the daughter product in part a decays by alpha emission and is in a high energy state.

c. Step 3: the high energy daughter product in part b decays by beta and gamma emissions and is in a high energy state.

d. Step 4: the daughter product in part c decays by beta and gamma emissions and is in a high energy state.

e. Step 5: the high energy daughter product in part d decays by beta emission.

f. Step 6: the daughter product in part e decays by alpha emission and is in a high energy state.

g. Step 7: the high energy daughter product in part f decays by beta and gamma emissions and is in a high energy state.

h. Step 8: the high energy daughter product in part g decays by alpha and gamma emissions.

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    $\begingroup$ Did you forget to type out part a) of the question? According to en.wikipedia.org/wiki/…, it should be an alpha decay. $\endgroup$ – TAR86 Feb 13 at 12:36
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    $\begingroup$ My bad. Part 1 a. Step 1: Radon-222 decays by alpha emission. (Radon has the element symbol Rn.) yes it is an alpha decay. I just dont understand how to explain all this through equation as this is literally my first problem like this ever. $\endgroup$ – Stu Pedaso Feb 13 at 12:39
  • $\begingroup$ Im asusming it starts as 222m/86 Rn----->0/-1 B+(gamma symbol) $\endgroup$ – Stu Pedaso Feb 13 at 12:44
  • $\begingroup$ You seem to be missing the essential difference between alpha and beta. $\endgroup$ – Ivan Neretin Feb 13 at 13:02
  • $\begingroup$ Believe it or not, that last comment helped A LOT. Thanks $\endgroup$ – Stu Pedaso Feb 13 at 13:11
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In any nuclear reactions the sum of the mass numbers $A_i$ (the sum of the indices on the top left) of the reagents and products is always the same. The same applies to nuclear charges $Z_i$ (indices in the lower left, which are often not indicated). Also, keep in mind the notation for metastable state:

$$ \begin{align} \ce{^{222}_{86}Rn &→ ^{218}_{84}Po + ^{4}_{2}He}\tag{a} \\ \ce{^{218}_{84}Po &→ ^{214\mathrm{m}}_{82}Pb + ^{4}_{2}He}\tag{b} \\ \ce{^{214\mathrm{m}}_{82}Pb &→ ^{214\mathrm{m}}_{83}Bi + ^{0}_{-1}e + γ}\tag{c} \\ \ce{^{214\mathrm{m}}_{83}Bi &→ ^{214\mathrm{m}}_{84}Po + ^{0}_{-1}e + γ}\tag{d} \\ \ce{^{214\mathrm{m}}_{84}Po &→ ^{214}_{85}At + ^{0}_{-1}e}\tag{e} \\ \ce{^{214}_{85}At &→ ^{210\mathrm{m}}_{83}Bi + ^{4}_{2}He}\tag{f} \\ \ce{^{210\mathrm{m}}_{83}Bi &→ ^{210\mathrm{m}}_{84}Po + ^{0}_{-1}e + γ}\tag{g} \\ \ce{^{210\mathrm{m}}_{84}Po &→ ^{206}_{82}Pb + ^{4}_{2}He + γ}\tag{h} \\ \end{align} $$

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    $\begingroup$ "High energy state" indicates that the nucleus will first decay with emission of a gamma ray, then a particle emission. $$\ce{^{210m}_{84}Po -> ^{210}_{84}Po + \gamma -> ^{206}_82Pb + ^4_2He }$$ $\endgroup$ – MaxW Feb 13 at 15:33
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    $\begingroup$ @MaxW That's right, but OP was asked to right down the equations for each "step", and those didn't differentiate gamma emission from the decay acts, unfortunately. Thanks for your comment though! $\endgroup$ – andselisk Feb 13 at 15:39
  • $\begingroup$ I understand this is a site where there are some people who are very serious about Chemistry. TBH this is my fourth week of this ever. I am also on Calculus one rn and I'm definitely struggling. I plan to explore as much chemistry as I can take. Im a veteran and a total meat head lol. It WILL take some getting used to. However, You guys are pretty awesome. I hope to exceed and study more sciences in the future and I appreciate any help I can get. $\endgroup$ – Stu Pedaso Feb 13 at 15:44
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    $\begingroup$ @andselisk: Step (h) in your answer should be: $\ce{^{210\mathrm{m}}_{84}Po -> ^{206}_{82}Pb + ^{4}_{2}He + \gamma}$ (not $\ce{^{210\mathrm{m}}_{84}Po -> ^{206}_{82}Po + ^{4}_{2}He + \gamma}$ as stated). :-) $\endgroup$ – Mathew Mahindaratne Feb 14 at 0:16
  • $\begingroup$ @MathewMahindaratne Thank you very much, fixed. I knew I would miss something as being lazy I iteratively copy-pasted equations:) $\endgroup$ – andselisk Feb 14 at 0:25

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