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

1 Answer

Answer:

#""_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:

  1. #""_92^234"U" → ""_90^230"Th" + ""_2^4"He"#
  2. #""_90^230"Th" → ""_88^226"Ra" + ""_2^4"He"#
  3. #""_88^226"Ra" → ""_86^222"Rn" + ""_2^4"He"#
  4. #""_86^222"Rn" → ""_84^218"Po" + ""_2^4"He"#
  5. #""_84^218"Po" → ""_82^214"Pb" + ""_2^4"He"#
  6. #""_82^214"Pb" → ""_83^214"Bi" + ""_text(-1)^0"e"#
  7. #""_83^214"Bi" → ""_84^214"Po" + ""_text(-1)^0"e"#
  8. #""_84^214"Po" → ""_82^210"Pb" + ""_2^4"He"#
  9. #""_82^210"Pb" → ""_83^210"Bi" + ""_text(-1)^0"e"#
  10. #""_83^210"Bi" → ""_84^210"Po" + ""_text(-1)^0"e"#
  11. #""_84^210"Po" → ""_82^206"Pb" + ""_2^4"He"#

Add them all up, and you get the overall equation

#""_92^234"U" → ""_82^206"Pb" + 7color(white)(l)""_2^4"He" + 4color(white)(l)""_text(-1)^0"e"#