Question:
If $^{}_{92}U^{206}$ emits 8 $\alpha$-particles and 6 $\beta$-particles, then the resulting nucleus is
If $^{}_{92}U^{206}$ emits 8 $\alpha$-particles and 6 $\beta$-particles, then the resulting nucleus is
Updated On: Jun 7, 2022
- $^{}_{82}U^{206}$
- $^{}_{82}Pb^{206}$
- $^{}_{82}U^{210}$
- $^{}_{82}U^{214}$
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The Correct Option is B
Solution and Explanation
After one $\alpha$-emission, the daughter nucleus reduces in mass number by 4 unit and in atomic number by 2 unit. In $\beta$-emission the atomic number of daughter nucleus increases by I unit. The reaction can be written as $^{}_{92}U^{238} \, \, \xrightarrow {- 8 \alpha} ^{}_{76}X^{206} \, \, \xrightarrow {- 6 \beta} \, \, ^{}_{82}Y^{206}$
Thus, the resulting nucleus is $^{}_{82}Y^{206}, \, ie , \, ^{}_{82}Pb^{206}$
Thus, the resulting nucleus is $^{}_{82}Y^{206}, \, ie , \, ^{}_{82}Pb^{206}$
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