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In the case of thorium $\left( {A = 232\;{\text{and}}\;Z = 90} \right)$ , we obtain an isotopes of lead $\left( {A = 208\;{\text{and}}\;Z = 82} \right)$ after some radioactive disintegrations. The number of $\alpha$ and $\beta -$ particles emitted are, respectively(A) $6,3$ (B) $6,4$(C) $5,5$(D) $4,6$

Last updated date: 19th Sep 2024
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Hint:- The alpha particle consists of two protons and two neutrons. Thus the mass number decreases by four when an alpha particle is emitted. And when the bet particle is emitted the number of protons will increase by one.

Step by step solution:
Given when the Thorium undergoes radioactive disintegration an isotope of Lead is obtained. The nuclear reaction can be expressed as
$_{90}T{h^{232}}{ \to _{82}}P{b^{208}} + {n_1}\alpha + {n_2}\beta -$
Where, ${n_1}$is the number of $\alpha$ particles and ${n_2}$ is the number of $\beta -$ particles.
The alpha article has mass number four. That is two protons and two neutrons. When an alpha particle is emitted the mass number of the parent nucleus will decrease by four. And while emitting the beta particle or gamma ray there is no change in the mass number.
We have given the mass number of the parent nucleus is $232$ and the mass number of isotope of lead is $208$ . Therefore the change happened in the mass number can be represented as,
$232 = 208 + 4{n_1}$
The number of alpha particles is multiplied by four, since the mass number of each alpha particle is four.
Therefore,
$4{n_1} = 232 - 208 \\ 4{n_1} = 24 \\ {n_1} = 6 \\$
Thus the number of alpha particles is $6$ .
The number of protons is equal to the atomic number. We have given the atomic number of Thorium as $90$ and the atomic number of Lead as $82$ . And for alpha particles the number of protons is two. And in the case of $\beta -$ decay the number of protons will be increased by one. The relation connecting the number of protons in the disintegration is given as,
$90 = 82 + 2{n_1} - {n_2}$
Substitute the value for ${n_1}$.
$90 = 82 + 2 \times 6 - {n_2} \\ {n_2} = 94 - 90 \\ {n_2} = 4 \\$
Thus the number of $\beta -$ particles is $4$