In the equation: $_{90}^{234} T h \to$ $"_90^234Th ->$ $_{91}^{234} P a + X$ $"_91^234Pa + X$ , which particle is represented by X?

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Answer 1 · 2017-02-01T14:57:43

The particle is an electron $=_{- 1}^{0} e$ $=""_-1^0e$

$_{90}^{234} T h \to_{91}^{234} P a + X$ $""_90^234Th->""_91^234Pa+X$

We have conservation of the number of nucleons and charge number

For $X$ $X$ , the number of nucleons is $= 0$ $=0$ and the charge number is $= - 1$ $=-1$

$_{- 1}^{0} X$ $""_-1^0 X$

This particle is $β^{-}$ $beta^-$ which is an electron $_{- 1}^{0} e$ $""_-1^0e$

In the equation: 23490Th→"_90^234Th -> 23491Pa+X"_91^234Pa + X, which particle is represented by X?