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Physics Question on Nuclei

If 92U206^{}_{92}U^{206} emits 8 α\alpha-particles and 6 β\beta-particles, then the resulting nucleus is

A

82U206^{}_{82}U^{206}

B

82Pb206^{}_{82}Pb^{206}

C

82U210^{}_{82}U^{210}

D

82U214^{}_{82}U^{214}

Answer

82Pb206^{}_{82}Pb^{206}

Explanation

Solution

After one α\alpha-emission, the daughter nucleus reduces in mass number by 4 unit and in atomic number by 2 unit. In β\beta-emission the atomic number of daughter nucleus increases by I unit. The reaction can be written as 92U238  →−8α76X206  →−6β  82Y206^{}_{92}U^{238} \, \, \xrightarrow {- 8 \alpha} ^{}_{76}X^{206} \, \, \xrightarrow {- 6 \beta} \, \, ^{}_{82}Y^{206}
Thus, the resulting nucleus is 82Y206, ie, 82Pb206^{}_{82}Y^{206}, \, ie , \, ^{}_{82}Pb^{206}