Question

2. The istope Sb may undergo both β and β+ decay. For both of these decay modes: (a) Write out the full reactions and calculate their respective Q values. Assume the daughter nuclei are in their respective ground states. (2 points) (b) Plot the normalized emission spectra p(T) for the electron kinetic energy on the plot, labelling the axes and providing a legend. Assume the Fermi function is Ze2 c F(Z, T,) = 1-exp(-2nr). where u is the particle speed (assume the particle is relativistic); see the lecture notes for the definition of the other symbols. Discuss the physical reasons for the differences between the β-and β+ spectra. (4 points)

Answer 1

The coulomb correction enhances the probability of electron emission and decreases the probability of positron emission, especially at low energies. The Coulomb force loses its effect on the spectrum shape at high energies and spectrum approaches that computed without the correction.

Explanation: The Coulomb field accelerates the positive electron and decelerates the negative electron. Hence the positron spectrum has fewer slow particles and the electron spectrum has more slow particles than they would have in the absence of the Coulomb correction.
The coulomb correction enhances the probability of electron emission and decreases the probability of positron emission, especially at low energies. The Coulomb force loses its effect on the spectrum shape at high energies and spectrum approaches that computed without the correction. Explanation: The Coulomb field accelerates the positive electron and decelerates the negative electron. Hence the positron spectrum has fewer slow particles and the electron spectrum has more slow particles than they would have in the absence of the Coulomb correction.(i) beta^-1 emission we have for beta^- emission X rightarrow Y^A + e^0 + v^bar + Q_beta rightarrow m_p c^2 = m_2 c^2 + Q rightarrow Q_B = (M_p - M_d) c^2 for sb sb^125_51 rightarrow Te^125_52 + e^0 + v^bar + Q_P rightarrow Q_B = (M_sb - M_Te) c^2