A) \[8\alpha \]and\[4\beta \]
B) \[8\alpha \]and\[16\beta \]
C) \[4\alpha \]and\[2\beta \]
D) \[6\alpha \]and\[4\beta \]
Correct Answer: D
Solution :
\[_{90}T{{h}^{232}}{{\xrightarrow{{}}}_{82}}P{{b}^{208}}\] In the above disintegration. Change in atomic weight \[=232-208=24\]. We know that for the emission of one \[\alpha -\]particle the atomic weight is decreased by 4. \[\therefore \]number of a-particles emitted \[=\frac{24}{4}=6\] After the emission of \[6\,\,\alpha -\]particles, the change in atomic number\[=90-12=78\]\[(\because \,\,\alpha {{=}_{2}}H{{e}^{4}})\]. The increase in atomic number from \[78\] to\[82=4\] \[\therefore \]number of \[\beta -\]particles emitted\[=4\] \[(\because \,\,\beta =-e{}^\circ )\] Hence, \[6\,\,\,\alpha -\]particles and \[4\,\,\beta -\]particles are emitted.
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