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${ }_{92} \mathrm{U}^{235}$ undergoes successive disintegrations with the end product of ${ }_{82} \mathrm{~Pb}^{203}$. The number of $\alpha$ and $\beta$ particles emitted are
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The correct answer is:
$\alpha=8, \beta=6$
When an $\alpha$ - particle emits from a radioactive nuclei, then its mass number is decreased by 4 unit and atomic number is decreased by 2 unit.
When a $\beta$-particle emits from a radioactive nuclei, then its mass number is unaffected, whereas atomic number is increased by one unit.
Here, initial nuclei $={ }_{92} \mathrm{U}^{235}$
Final nuclei $={ }_{82} \mathrm{P}^{2013}$
Change in mass number $=235-203=32$
Number of $\alpha$-particle emitted $=\frac{32}{4}=8$
$\therefore$ Number of emitted $\beta$-particles
$$
=8 \times 2-(92-82)=16-10=6
$$
When a $\beta$-particle emits from a radioactive nuclei, then its mass number is unaffected, whereas atomic number is increased by one unit.
Here, initial nuclei $={ }_{92} \mathrm{U}^{235}$
Final nuclei $={ }_{82} \mathrm{P}^{2013}$
Change in mass number $=235-203=32$
Number of $\alpha$-particle emitted $=\frac{32}{4}=8$
$\therefore$ Number of emitted $\beta$-particles
$$
=8 \times 2-(92-82)=16-10=6
$$
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