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

a

\( ^0_1e \)

b

\( ^0_{-1}e \)

c

\( ^4_2He \)

d

\( ^1_0n \)

e

\( ^1_1p\)

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Correct Option
b

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Discussions (4)

adisco4real
2 years ago

B its coreect thats beta emission

Tega204
2 years ago

none of the above. X should be a positron +1⁰e that'll reduce the atomic number by 1.

Tega204
2 years ago

A is legit ✅. positron ⁰1 e or ⁰+1e

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