
If \[{}_{{\rm{92}}}{{\rm{U}}^{{\rm{238}}}}\] undergoes successively 8 \[{\alpha^- }\]decays and 6 \[{\beta ^ - }\]decays, then resulting nucleus is
A. \[{}_{{\rm{82}}}{{\rm{U}}^{{\rm{236}}}}\]
B. \[{}_{{\rm{82}}}{\rm{P}}{{\rm{b}}^{{\rm{206}}}}\]
C. \[{}_{{\rm{82}}}{{\rm{U}}^{{\rm{210}}}}\]
D. \[{}_{{\rm{82}}}{{\rm{U}}^{{\rm{214}}}}\]
Answer
162.3k+ views
Hint: In one alpha decay atomic number Z of daughter nuclei reduces by 2 units and mass number A is reduced by 4 units while in one beta negative decay atomic number increases by 1 unit and the mass number remains the same.
Complete answer:
Radioactive isotope\[{}_{{\rm{92}}}{{\rm{U}}^{{\rm{238}}}}\]of Uranium undergoes 8 \[{\alpha^- }\]decays and 6 \[{\beta ^ - }\]decays, we have to find the resulting nucleus after decay.
As we know that in one alpha decay atomic number reduces by 4 units and mass number is reduced by 2 units, therefore in 8\[{\alpha^- }\]decays,
Atomic number reduces by \[8 \times 2 = 16\,\,units\]
Mass number reduces by \[8 \times 4 = 32\,\,units\]
Therefore after 8 \[\alpha - \]decays we have,
\[{}_{{\rm{92}}}{{\rm{U}}^{{\rm{238}}}} \to {}_{{\rm{92 - 16}}}{{\rm{U}}^{{\rm{238 - 32}}}} = {}_{{\rm{76}}}{{\rm{X}}^{{\rm{206}}}}\]
And in one beta negative decay atomic number increases by 1 unit while the mass number remains same, therefore after 6 \[{\beta ^ - }\]decay of \[{}_{{\rm{76}}}{{\rm{X}}^{{\rm{206}}}}\]
\[{}_{{\rm{76}}}{{\rm{X}}^{{\rm{206}}}} \to {}_{{\rm{76 + 6}}}{{\rm{X}}^{{\rm{206}}}} = {}_{{\rm{82}}}{{\rm{X}}^{{\rm{206}}}}\]
As the resulting nucleus \[{}_{{\rm{82}}}{{\rm{X}}^{{\rm{206}}}}\] has mass number 82 which is for lead (Pb), therefore resulting nucleus will be \[{}_{{\rm{82}}}{\rm{P}}{{\rm{b}}^{{\rm{206}}}}\].
Hence, 8 \[{\alpha^- }\]decays and 6 \[{\beta ^ - }\]decay of \[{}_{{\rm{92}}}{{\rm{U}}^{{\rm{238}}}}\]results into \[{}_{{\rm{82}}}{\rm{P}}{{\rm{b}}^{{\rm{206}}}}\].
Therefore, option B is the correct option.
Note: In beta negative decay parent nuclei decays into daughter nuclei with emission of energetic electron particle and mass less antineutrino and as the antineutrino has negligible or no mass therefore it does not affect mass number or atomic number of the daughter nuclei.
Complete answer:
Radioactive isotope\[{}_{{\rm{92}}}{{\rm{U}}^{{\rm{238}}}}\]of Uranium undergoes 8 \[{\alpha^- }\]decays and 6 \[{\beta ^ - }\]decays, we have to find the resulting nucleus after decay.
As we know that in one alpha decay atomic number reduces by 4 units and mass number is reduced by 2 units, therefore in 8\[{\alpha^- }\]decays,
Atomic number reduces by \[8 \times 2 = 16\,\,units\]
Mass number reduces by \[8 \times 4 = 32\,\,units\]
Therefore after 8 \[\alpha - \]decays we have,
\[{}_{{\rm{92}}}{{\rm{U}}^{{\rm{238}}}} \to {}_{{\rm{92 - 16}}}{{\rm{U}}^{{\rm{238 - 32}}}} = {}_{{\rm{76}}}{{\rm{X}}^{{\rm{206}}}}\]
And in one beta negative decay atomic number increases by 1 unit while the mass number remains same, therefore after 6 \[{\beta ^ - }\]decay of \[{}_{{\rm{76}}}{{\rm{X}}^{{\rm{206}}}}\]
\[{}_{{\rm{76}}}{{\rm{X}}^{{\rm{206}}}} \to {}_{{\rm{76 + 6}}}{{\rm{X}}^{{\rm{206}}}} = {}_{{\rm{82}}}{{\rm{X}}^{{\rm{206}}}}\]
As the resulting nucleus \[{}_{{\rm{82}}}{{\rm{X}}^{{\rm{206}}}}\] has mass number 82 which is for lead (Pb), therefore resulting nucleus will be \[{}_{{\rm{82}}}{\rm{P}}{{\rm{b}}^{{\rm{206}}}}\].
Hence, 8 \[{\alpha^- }\]decays and 6 \[{\beta ^ - }\]decay of \[{}_{{\rm{92}}}{{\rm{U}}^{{\rm{238}}}}\]results into \[{}_{{\rm{82}}}{\rm{P}}{{\rm{b}}^{{\rm{206}}}}\].
Therefore, option B is the correct option.
Note: In beta negative decay parent nuclei decays into daughter nuclei with emission of energetic electron particle and mass less antineutrino and as the antineutrino has negligible or no mass therefore it does not affect mass number or atomic number of the daughter nuclei.
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