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Answer 1 · 2017-05-02T17:15:24

$_{82}^{211} Pb \to_{83}^{211} Bi +_{-1}^{0} e$ $""_82^211"Pb" → ""_83^211"Bi" + ""_text(-1)^0"e"$

It's difficult to read your question. I assume that you want to balance an equation like this:

$_{82}^{211} Pb \to_{83}^{211} Bi +_{?}^{?} ?$ $""_82^211"Pb" → ""_83^211"Bi" + ""_text(?)^??$

The main things to remember are:

Subscripts

$82 = 83 + ?$ $82 = 83 + ?$
$? = 82 - 83 = -1$ $? = "82 - 83" = "-1"$

This gives

$_{82}^{211} Pb \to_{83}^{211} Bi +_{-1}^{?} ?$ $""_82^211"Pb" → ""_83^211"Bi" + ""_text(-1)^??$

Superscripts

$211 = 211 + ?$ $211 = 211 + ?$
$? = 211 - 211 = 0$ $? = "211 - 211" = 0$

This gives

$_{82}^{211} Pb \to_{83}^{211} Bi +_{-1}^{0} ?$ $""_82^211"Pb" → ""_83^211"Bi" + ""_text(-1)^0?$

Identify the particle

What particle has a charge of -1 and a mass near zero? An electron (or beta particle)!

The complete equation becomes

$_{82}^{211} Pb \to_{83}^{211} Bi +_{-1}^{0} e$ $""_82^211"Pb" → ""_83^211"Bi" + ""_text(-1)^0"e"$