Question

Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme:
¹²N → ¹²C* + e⁺ + v
¹²C* → ¹²C + γ (4.43MeV).
The atomic mass of ¹²N is 12.018613 u.

Answer 1

Given:
Atomic mass of ¹²N, m(¹²N) = 12.018613 u
¹²N → ¹²C* + e⁺ + v
¹²C* → ¹²C + γ (4.43 MeV)

Net reaction is given by
¹²N → ¹²C + e⁺ + v + γ (4.43 MeV)

Q_value of the ${\beta }^{+}$ decay will be
Q_value= [ m(¹²N) $-$ (m(¹²C*) + 2m_e)]c²
= [12.018613 $\times$931 MeV $-$ (12$\times$931 + 4.43) MeV $-$ (2$\times$511) keV]
= [11189.3287 $-$ 11176.43 $-$ 1.022] MeV
= 11.8767 MeV = 11.88 MeV

The maximum kinetic energy of beta particle will be 11.88 MeV, assuming that neutrinos have zero energy.