We study a process called double beta^- decay, in which 2 electrons are emitted

Question

We study a process called double beta^- decay, in which 2 electrons are emitted simultaneously, in order to determine if the neutrino is a Majorana particle, that is, if it is its own antiparticle. We first consider the dominant process (called 2 beta 2v) A_Z X _N rightarrow A_Z' Y_N' + 2e^- + 2v_e where A = N + Z = N' + Z' is the total number of nucleons, N the number of neutrons of X and Z the number of protons of X. What are the numbers of protons Z' and neutrons N' of the daughter nucleus Y as a function of Z and N? ^76 Ge (Z = 32) is one of the few nuclei which encounter double beta^- decay but not single beta^- decay. Using the properties of nuclei at the end of this question, calculate the Q-value for the single beta^- decay and for the double beta^- decay. Show that, indeed, only the double beta^- decay is energetically possible. We now consider the process (called 2 beta 0v) A_Z X_N rightarrow A_Z'Y_N' + 2e^-, where only the two electrons are emitted in the final state. This process is possible only if the neutrino is a Majorana particle. Note that the kinetic energy of the daughter nucleus Y is negligible with respect to the kinetic energy of the electrons. Imagine that you have a detection setup which allows you to measure the kinetic energy T_e1 + T_e2 of both electrons simultaneously, and that both processes (2) and (3) are happening. Sketch the distribution of T_e1 + T_e2 (number of events in the vertical axis and total kinetic energy of both electrons in the horizontal axis). Arsenic (As) has Z = 33 protons and Selenium (Se) has Z = 34 protons. The difference of nuclear masses of their A = 76 isotopes with respect to ^76 Ge are: M(^76 Ge) - M(^76 Se) asymptoticallyequalto 2.039 MeV/c^2 M(^76 Ge) - M(^76 As) asymptoticallyequalto -0.922 MeV/c^2. The mass of the electron is m_e asymptoticallyequalto 0.511 MeV/c^2. The neutrinos are assumed to be massless.

Explanation / Answer

Q2-1. Here the equation for Z in terms of Z/ according to conservation of charge number is as

Z = Z/ + 2(-1) + 2(0), (since electron has charge number = -1 and antineutrino has charge number = 0)

Thus, Z/ = Z + 2................................(1)

Now, N+Z = N/+Z/

Or, N+Z = N/+Z+2 [from equation(1)]

Thus, N/ = N+2................................(2)

Equations (1) and (2) give the required expressions.   

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