What is the net reaction for the nuclear decay chain of $U - 234$ $"U"-234$ ?

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What is the net reaction for the nuclear decay chain of $U - 234$ $"U"-234$ ?

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Answer 1 · 2017-07-25T16:05:46

$_{92}^{234} U \to_{82}^{206} Pb + 7 l_{2}^{4} He + 4 l_{-1}^{0} e$ $""_92^234"U" → ""_82^206"Pb" + 7color(white)(l)""_2^4"He" + 4color(white)(l)""_text(-1)^0"e"$

Explanation:

Uranium-234 decays to lead-206 in a series of 11 steps involving seven α-decays and four β-decays.

Here are the steps:

$_{92}^{234} U \to_{90}^{230} Th +_{2}^{4} He$ $""_92^234"U" → ""_90^230"Th" + ""_2^4"He"$
$_{90}^{230} Th \to_{88}^{226} Ra +_{2}^{4} He$ $""_90^230"Th" → ""_88^226"Ra" + ""_2^4"He"$
$_{88}^{226} Ra \to_{86}^{222} Rn +_{2}^{4} He$ $""_88^226"Ra" → ""_86^222"Rn" + ""_2^4"He"$
$_{86}^{222} Rn \to_{84}^{218} Po +_{2}^{4} He$ $""_86^222"Rn" → ""_84^218"Po" + ""_2^4"He"$
$_{84}^{218} Po \to_{82}^{214} Pb +_{2}^{4} He$ $""_84^218"Po" → ""_82^214"Pb" + ""_2^4"He"$
$_{82}^{214} Pb \to_{83}^{214} Bi +_{-1}^{0} e$ $""_82^214"Pb" → ""_83^214"Bi" + ""_text(-1)^0"e"$
$_{83}^{214} Bi \to_{84}^{214} Po +_{-1}^{0} e$ $""_83^214"Bi" → ""_84^214"Po" + ""_text(-1)^0"e"$
$_{84}^{214} Po \to_{82}^{210} Pb +_{2}^{4} He$ $""_84^214"Po" → ""_82^210"Pb" + ""_2^4"He"$
$_{82}^{210} Pb \to_{83}^{210} Bi +_{-1}^{0} e$ $""_82^210"Pb" → ""_83^210"Bi" + ""_text(-1)^0"e"$
$_{83}^{210} Bi \to_{84}^{210} Po +_{-1}^{0} e$ $""_83^210"Bi" → ""_84^210"Po" + ""_text(-1)^0"e"$
$_{84}^{210} Po \to_{82}^{206} Pb +_{2}^{4} He$ $""_84^210"Po" → ""_82^206"Pb" + ""_2^4"He"$

Add them all up, and you get the overall equation

$_{92}^{234} U \to_{82}^{206} Pb + 7 l_{2}^{4} He + 4 l_{-1}^{0} e$ $""_92^234"U" → ""_82^206"Pb" + 7color(white)(l)""_2^4"He" + 4color(white)(l)""_text(-1)^0"e"$

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What is the net reaction for the nuclear decay chain of $U - 234$ $"U"-234$ ?

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