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.

(Use Mass of proton m_p = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron m_n = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c²,1 u = 931 MeV/c².)

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 `β^+` decay will be

Q_value = [m(`""^12N`) - (m(¹²C*) + 2m_e)]c²

`= [12.018613 xx 931 "MeV" - (12 xx 931 + 4.43) "MeV" - (2 xx 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.