Nuclear equations for the decay of radon-222 to lead-206 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.
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:

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