When the radioactive isotope $_{88}{Ra}^{226}$ decays in a series by emission of three alpha ($\alpha$) and a beta ($\beta$) particle, the isotope X which remains undecayed is?
A. $_{83}{X}^{214}$
B. $_{84}{X}^{218}$
C. $_{84}{X}^{220}$
D. $_{87}{X}^{223}$
Answer
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Hint: To solve this problem, first write the alpha decay of $_{88}{Ra}^{226}$ where $_{88}{Ra}^{226}$ is the parent nucleus. Then, write the alpha decay for the daughter nucleus obtained through earlier decay. Again, write the alpha decay for this obtained daughter nucleus. These three decays show that three alpha particles are emitted. Then, use the obtained daughter nucleus to write the beta decay. In this way three alpha and a beta particle will be emitted in series. The daughter nucleus obtained does not decay further. So, it is the element which remains undecayed.
Complete answer:
Alpha decay is given by,
$ _{Z}{X}^{A} \rightarrow _{Z-2}{Y}^{A-4} + _{2}{\alpha}^{4}$
After first alpha emission, $_{88}{Ra}^{226}$ decays as given below,
$ _{88}{Ra}^{226} \rightarrow _{86}{Rn}^{222} + _{2}{\alpha}^{4}$
Now, the second alpha emission will be,
$ _{86}{Rn}^{222} \rightarrow _{84}{Po}^{218} + _{2}{\alpha}^{4}$
After the third alpha emission,
$ _{84}{Po}^{218} \rightarrow _{82}{Pb}^{214} + _{2}{\alpha}^{4}$
Beta decay is given by,
$ _{Z}{X}^{A} \rightarrow _{Z+1}{Y}^{A} + {\beta}^{-}$
After three alpha emissions, a beta particle is emitted. So, the decay can be given by,
$ _{82}{Pb}^{214} \rightarrow _{83}{Bi}^{214} + {\beta}^{-}$
After a series emission of three alpha and a beta particle, the decay can be given as,
$ _{88}{Ra}^{226} \rightarrow _{83}{Bi}^{214} + 3_{2}{\alpha}^{4} + {\beta}^{-}$
So, after series by emission of three alpha ($\alpha$) and a beta ($\beta$) particle, the isotope X which remains undecayed is $_{83}{X}^{214}$.
So, the correct answer is “Option A”.
Note:
In practice, alpha decay is only observed in the nuclides considerably heavier than nickel. Alpha particles are emitted by all the heavy radioactive nuclides occurring in nature. Examples of these heavy radioactive nuclides include thorium, uranium, etc. Alpha particles are also emitted by transuranic elements like neptunium, americium, etc. After an alpha decay, the daughter nucleus is often left in an excited energy state. Beta decay is relatively a slow process as compared to other decay processes.
Complete answer:
Alpha decay is given by,
$ _{Z}{X}^{A} \rightarrow _{Z-2}{Y}^{A-4} + _{2}{\alpha}^{4}$
After first alpha emission, $_{88}{Ra}^{226}$ decays as given below,
$ _{88}{Ra}^{226} \rightarrow _{86}{Rn}^{222} + _{2}{\alpha}^{4}$
Now, the second alpha emission will be,
$ _{86}{Rn}^{222} \rightarrow _{84}{Po}^{218} + _{2}{\alpha}^{4}$
After the third alpha emission,
$ _{84}{Po}^{218} \rightarrow _{82}{Pb}^{214} + _{2}{\alpha}^{4}$
Beta decay is given by,
$ _{Z}{X}^{A} \rightarrow _{Z+1}{Y}^{A} + {\beta}^{-}$
After three alpha emissions, a beta particle is emitted. So, the decay can be given by,
$ _{82}{Pb}^{214} \rightarrow _{83}{Bi}^{214} + {\beta}^{-}$
After a series emission of three alpha and a beta particle, the decay can be given as,
$ _{88}{Ra}^{226} \rightarrow _{83}{Bi}^{214} + 3_{2}{\alpha}^{4} + {\beta}^{-}$
So, after series by emission of three alpha ($\alpha$) and a beta ($\beta$) particle, the isotope X which remains undecayed is $_{83}{X}^{214}$.
So, the correct answer is “Option A”.
Note:
In practice, alpha decay is only observed in the nuclides considerably heavier than nickel. Alpha particles are emitted by all the heavy radioactive nuclides occurring in nature. Examples of these heavy radioactive nuclides include thorium, uranium, etc. Alpha particles are also emitted by transuranic elements like neptunium, americium, etc. After an alpha decay, the daughter nucleus is often left in an excited energy state. Beta decay is relatively a slow process as compared to other decay processes.
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