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  4. When 90 Th 228 transforms to 83 Bi 212, then the number to the emitted α and β-particles are, respectively

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Q. When ${}_{90} Th ^{228}$ transforms to ${}_{83} Bi ^{212}$, then the number to the emitted $\alpha$ and $\beta$-particles are, respectively

NTA AbhyasNTA Abhyas 2022

Solution:

$Z={}_{90} T h^{A=228} \rightarrow={}_{83} B i^{A^{\prime}=212} $
Number of $\alpha$-particles emitted
$n_{\alpha}=\frac{A-A^{\prime}}{4}=\frac{228-212}{4}=4$
Number of $\beta$-particles emitted
$ n_{\beta}=2 n_{\alpha}-Z+Z^{\prime}=2 \times 4-90+83=1 \text {. }$