Question

The radioactive decay of ${}_5^{12}\text{B}$ to produce ${}_6^{12}\text{C}$ with 13.4 MeV energy may involve either of the way:

• A. ${}_5^{12}\text{B}$ decays directly to stable ${}_6^{12}\text{C}$ by emitting 13.4 MeV electron
• B. ${}_5^{12}\text{B}$ decays to excited ${}_6^{12}\text{C}$ by emitting 9.0 MeV electron followed by decay of excited ${}_6^{12}\text{C}$ to give stable ${}_6^{12}\text{C}$ and a 4.4 MeVphoton
• C. ${}_5^{12}\text{B}{⟶}_6^{12}{\text{C}}^{\ast }+1_{-1}^0\text{e}+\overline{v}+13.4MeV$
• D. ${}_6^{12}{\text{C}}^{\ast }{⟶}_6^{12}\text{C}{+}_{-1}^0\text{e}+\overline{v}+4.4MeV$

Answer 1

Answer: (A), (B)

${}_5^{12}\text{B}$ decays to excited ${}_6^{12}\text{C}$ by emitting 9.0 MeV electron followed by decay of excited ${}_6^{12}\text{C}$ to give stable ${}_6^{12}\text{C}$ and a 4.4 MeVphoton
${}_5^{12}\text{B}$ decays directly to stable ${}_6^{12}\text{C}$ by emitting 13.4 MeV electron

In radioactive decay, it may be case that, direct ground state daughter element forms, also its a possibility that a excited state element forms which further looses energy and forms stable ground state element.