Question

Carbon nucleus $_6C^{11}$ decays into boron $_5B^{11}$ by $\beta^+$ decay. Determine Q-value of this process (provided atomic mass of carbon = 20073 $m_e$; atomic mass of boron = 20060 $m_e$; $m_e$ is electronic mass)

• A. 13 $m_e$ $C^2$
• B. 12 $m_e$ $C^2$
• C. 11 $m_e$ $C^2$
• D. 14 $m_e$ $C^2$

Answer 1

Answer: (C)

11 $m_e$ $C^2$

Answer 2

The Q-value of a nuclear decay is the energy released during the process, calculated from the mass difference between the reactants and products. For $\beta^+$ decay, the reaction is: $_Z^A X \rightarrow _ {Z-1}^A Y + e^+ + \nu_e$.

When using atomic masses ($M_{atom}$), the Q-value for $\beta^+$ decay is given by the formula: $Q = [ M_{atom}(X) - M_{atom}(Y) - 2m_e ] c^2$ where $M_{atom}(X)$ is the atomic mass of the parent nucleus, $M_{atom}(Y)$ is the atomic mass of the daughter nucleus, and $m_e$ is the mass of an electron (which is equal to the mass of a positron). The factor of $2m_e$ arises from accounting for the electron masses associated with the atomic masses and the emitted positron.

Given: Atomic mass of carbon $_6^{11}C$ ($M_{atom}(X)$) = $20073 m_e$ Atomic mass of boron $_5^{11}B$ ($M_{atom}(Y)$) = $20060 m_e$

Substituting these values into the Q-value formula: $Q = [ 20073 m_e - 20060 m_e - 2 m_e ] c^2$ $Q = [ (20073 - 20060 - 2) m_e ] c^2$ $Q = [ 13 m_e - 2 m_e ] c^2$ $Q = 11 m_e c^2$

The Q-value of this process is $11 m_e c^2$.