- #1
says
- 594
- 12
Homework Statement
1) Calculate the Q-value for the electron emission beta decay of Co-60 *
2) Calculate the Q-value for the positron emission beta decay of Mg-23 *
* (both correct to 6 significant figures)
Atomic Masses (u)
electron = 0.0005485803
Co-60 = 59.9338222
Ni-60 = 59.9307906
Mg-23 = 22.9941249
Na-23 = 22.98976967
Atomic Number
Co-27
Ni-28
Mg-12
Na-11
Homework Equations
Q = Σmic2-Σmfc2
Beta decay (electron emission) = X ---> Y(Z+1) + -1e + v_
where v_ = anti-neutrino
Beta decay (positron emission) = X ---> Y(Z-1) + +1e + v
where v = neutrino
The Attempt at a Solution
1 (Co-60)
Q = (59.9338222 - (59.9307906+0.0005485803))*c2
= 59.9338222 - 59.9313392
= 0.0024830 * 931.5020
= 2.3129
= 2.31290 MeV (correct to 6 significant figures)
2 (Mg-23)
Q = (22.9941249 - (22.98976967+0.0005485803))*c2
= 22.9941249-22.99031825
= 0.0038066*931.5020
= 3.54580 MeV (correct to 6 significant figures)
Q is positive in both cases. Q is the energy that is liberated in the reaction. Am i correct in neglecting and mass that the anti-neutrino and neutrino have?