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.

closed as off-topic by Jon Custer, Tyberius, Nuclear Chemist, A.K., Todd MinehardtFeb 16 at 1:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

• Did you forget to type out part a) of the question? According to en.wikipedia.org/wiki/…, it should be an alpha decay. – TAR86 Feb 13 at 12:36
• 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. – Stu Pedaso Feb 13 at 12:39
• Im asusming it starts as 222m/86 Rn----->0/-1 B+(gamma symbol) – Stu Pedaso Feb 13 at 12:44
• You seem to be missing the essential difference between alpha and beta. – Ivan Neretin Feb 13 at 13:02
• Believe it or not, that last comment helped A LOT. Thanks – Stu Pedaso Feb 13 at 13:11

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}
• "High energy state" indicates that the nucleus will first decay with emission of a gamma ray, then a particle emission. $$\ce{^{210m}_{84}Po -> ^{210}_{84}Po + \gamma -> ^{206}_82Pb + ^4_2He }$$ – MaxW Feb 13 at 15:33
• @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). :-) – Mathew Mahindaratne Feb 14 at 0:16