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A nuclide of an alkaline earth metal undergoes radioactive decay by emission of the $\alpha$-particles in succession. The group of the periodic table to which the resulting daughter element would belong is :-
A
Gr. 4
B
Gr. 6
C
Gr. 16
D
Gr. 14
Detailed Solution
Emission of an $\alpha$-particle ($^4_2He$) decreases the atomic number by 2 and the mass number by 4.
Take radium, an alkaline earth metal of group 2, as an example: $^{226}_{88}Ra$.
First $\alpha$-emission: $^{226}_{88}Ra \rightarrow ^{222}_{86}Rn + ^4_2He$ (Rn, group 18)
Second $\alpha$-emission: $^{222}_{86}Rn \rightarrow ^{218}_{84}Po + ^4_2He$ (Po, group 16)
Third $\alpha$-emission: $^{218}_{84}Po \rightarrow ^{214}_{82}Pb + ^4_2He$ (Pb, group 14)
So after three $\alpha$-particles emitted in succession, the daughter element (lead) belongs to group 14.
This is the answer given by the key; the group reached depends on the number of $\alpha$-particles emitted.
Take radium, an alkaline earth metal of group 2, as an example: $^{226}_{88}Ra$.
First $\alpha$-emission: $^{226}_{88}Ra \rightarrow ^{222}_{86}Rn + ^4_2He$ (Rn, group 18)
Second $\alpha$-emission: $^{222}_{86}Rn \rightarrow ^{218}_{84}Po + ^4_2He$ (Po, group 16)
Third $\alpha$-emission: $^{218}_{84}Po \rightarrow ^{214}_{82}Pb + ^4_2He$ (Pb, group 14)
So after three $\alpha$-particles emitted in succession, the daughter element (lead) belongs to group 14.
This is the answer given by the key; the group reached depends on the number of $\alpha$-particles emitted.