\( ^{234}_{90}Th → ^{234}_{91}Pa + X \) In the nuclear reaction above, X is

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Chemistry NECO 2008

\( ^{234}_{90}Th → ^{234}_{91}Pa + X \)
In the nuclear reaction above, X is

  • \( ^0_1e \)
  • \( ^0_{-1}e \) checkmark
  • \( ^4_2He \)
  • \( ^1_0n \)
  • \( ^1_1p\)

The correct answer is: B

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