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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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Verified Answer
The correct answer is:
$\alpha=8, \beta=6$
Let number of $\alpha$ particles decayed be $x$ and number of $\beta$ particles decayed be $y$.
Then equation for the decay is given by
$$
{ }_{92} \mathrm{U}^{235} \longrightarrow \mathrm{x} \alpha_{2}^{4}+\mathrm{y} \beta_{-1}^{0}+\mathrm{Pb}_{82}^{203}
$$
Equating the mass number on both sides
$$
235=4 x+203
$$
Equating atomic number on both sides
$$
92=2 \mathrm{x}-\mathrm{y}+82
$$
Solving Eqs. (i) and (ii), we get
$$
x=8, y=6
$$
$\therefore 8 \alpha$ particles and $6 \beta$ particles are emitted in disintegration.
Then equation for the decay is given by
$$
{ }_{92} \mathrm{U}^{235} \longrightarrow \mathrm{x} \alpha_{2}^{4}+\mathrm{y} \beta_{-1}^{0}+\mathrm{Pb}_{82}^{203}
$$
Equating the mass number on both sides
$$
235=4 x+203
$$
Equating atomic number on both sides
$$
92=2 \mathrm{x}-\mathrm{y}+82
$$
Solving Eqs. (i) and (ii), we get
$$
x=8, y=6
$$
$\therefore 8 \alpha$ particles and $6 \beta$ particles are emitted in disintegration.
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