- #1
oleosquarewave
- 7
- 4
- Homework Statement
- The nucleus ##^{198}\text{Hg}## has excited states at ##0.412 \, MeV## and ##1.088 \, MeV##. Following the beta decay of ##^{198}\text{Au}## to ##^{198}\text{Hg}##, three gamma rays are emitted. Find the energies of these three gamma rays.
- Relevant Equations
- ##Q=[m(^AX)-m(^AX')]c^2##
This is what I have for my solution thus far:
"For gold to transition to mercury, it must exchange one of it's neutrons for a proton via ##\beta##-decay. In order to find the energy of the emitted gamma rays, we must first find the excess rest energy ##Q## present after the ##\beta##-decay. The neutron decay process will emit an electron and an antineutrino.
$$
^{198}_{79}\text{Au}_{119}\to ^{198}_{80}\text{Hg}_{118}+e^{-}+\bar{\nu}
$$
This process will have a ##Q## value of
$$
Q=[m(^{198}\text{Au})-m(^{198}\text{Hg})]c^{2}=[197.968244 \, u-197.966769 \, ](931.50 \, MeV / u)=1.374 \, MeV
$$
This excess energy appears as the kinetic energy of the electron, the energy of the antineutrino, and a negligible kinetic energy from the recoil of the newly created mercury nucleus."
From here, I am a little lost on where to go. I know we are given the excited states for mercury-198, and my initial (and honestly probably correct) intuition is to just calculate the possible transition energies from excited states to lower-energy/ground states, but the textbook (Modern Physics Krane 4th Ed.) had this whole buildup of calculating the excess energy present after decay processes, and I know that ##\beta## decay from neutron to proton necessitates that the excess energy be realized as the kinetic energy of the electron and the energy of the antineutrino, which confuses the issue of how much energy these three gamma rays will have. Also, dumb question, but why three gamma rays?
"For gold to transition to mercury, it must exchange one of it's neutrons for a proton via ##\beta##-decay. In order to find the energy of the emitted gamma rays, we must first find the excess rest energy ##Q## present after the ##\beta##-decay. The neutron decay process will emit an electron and an antineutrino.
$$
^{198}_{79}\text{Au}_{119}\to ^{198}_{80}\text{Hg}_{118}+e^{-}+\bar{\nu}
$$
This process will have a ##Q## value of
$$
Q=[m(^{198}\text{Au})-m(^{198}\text{Hg})]c^{2}=[197.968244 \, u-197.966769 \, ](931.50 \, MeV / u)=1.374 \, MeV
$$
This excess energy appears as the kinetic energy of the electron, the energy of the antineutrino, and a negligible kinetic energy from the recoil of the newly created mercury nucleus."
From here, I am a little lost on where to go. I know we are given the excited states for mercury-198, and my initial (and honestly probably correct) intuition is to just calculate the possible transition energies from excited states to lower-energy/ground states, but the textbook (Modern Physics Krane 4th Ed.) had this whole buildup of calculating the excess energy present after decay processes, and I know that ##\beta## decay from neutron to proton necessitates that the excess energy be realized as the kinetic energy of the electron and the energy of the antineutrino, which confuses the issue of how much energy these three gamma rays will have. Also, dumb question, but why three gamma rays?