A radioactive element _{90}X^{238} decays int | vedclass

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(D) Let $n_{\alpha}$ be the number of $\alpha$-particles and $n_{\beta}$ be the number of $\beta$-particles emitted.The change in mass number is given

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$1$

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(D) Let $n_{\alpha}$ be the number of $\alpha$-particles and $n_{\beta}$ be the number of $\beta$-particles emitted.The change in mass number is given by: $238 - 222 = 4 	imes n_{\alpha}$.Thus,$n_{\alpha} = \frac{16}{4} = 4$.The change in atomic number is given by: $90 - 83 = 2 	imes n_{\alpha} - 1 	imes n_{\beta}$.Substituting $n_{\alpha} = 4$ into the equation:$7 = 2(4) - n_{\beta}$.$7 = 8 - n_{\beta}$.$n_{\beta} = 8 - 7 = 1$.Therefore,$1$ $\beta$-particle is emitted.

$A$ radioactive element $_{90}X^{238}$ decays into $_{83}Y^{222}$. The number of $\beta$-particles emitted is:

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