Question

Which ones of the following reactions are artificial radioactive decay?

• A. ${}_{92}{}^{238}U{\to }_{90}{}^{234}Th+\alpha$
• B. ${}_{12}{}^{24}Mg+{\beta }^{+}{\to }_{13}{}^{24}Al$
• C. ${}_4{}^{9}Be{+}_2{}^{4}He{\to }_6{}^{12}C+n$
• D. ${}_{92}{}^{238}U{\to }_{90}{}^{234}Th+\beta$

Answer 1

Answer: (B), (C)

${}_4{}^{9}Be{+}_2{}^{4}He{\to }_6{}^{12}C+n$

${}_{92}{}^{238}U{\to }_{90}{}^{234}Th+\alpha$
In this reaction ${}_{92}{}^{238}U$ does $\alpha$ decay and ${}_{90}{}^{234}Th$ is formed. It is natural decay.
${}_{12}{}^{24}Mg+{\beta }^{+}{\to }_{13}{}^{24}Al$
In this reaction nonradioactive Mg atom does ${\beta }^{+}$ and ${}_{13}{}^{24}Al$ is formed. It is artificial decay.
${}_4{}^{9}Be{+}_2{}^{4}He{\to }_6{}^{12}C+n$
In this reaction nonradioactive Be atom does ${}_2{}^{4}He$ decay and ${}_6{}^{12}C$ is formed. It is also artificial decay.