How to calculate gamma ray energies following beta decay?

  • #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?
 
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  • #2
The Q-value you have computed is for decay directly to the mercury ground state I assume?

oleosquarewave said:
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.
Are you perhaps forgetting something in this list?
 
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  • #3
oleosquarewave said:
Homework Statement: The nucleus ##^{198}\text{Hg}## has excited states at ##0.412 \, eV## and ....
Hopefully you have a typo' and mean ##0.412MeV##.

oleosquarewave said:
Following the beta decay of ##^{198}\text{Au}## to ##^{198}\text{Hg}##, three gamma rays are emitted.
The wording (IMO) implies that it's OK for you to assume that (at least some of) the decays leave the Hg-198 nuclei in the 1.08MeV excited state. If so, the Q-value is not relevant. Presumably the given energy-level values are based on taking the ground-state as zero.

(But you should still reflect on @Orodruin's comments on what you wrote.)

oleosquarewave said:
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,
Yes, that's what I'd do.

oleosquarewave said:
Also, dumb question, but why three gamma rays?
How many different levels are of interest? How many possible transitions are there between these levels?
 
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  • #4
Orodruin said:
The Q-value you have computed is for decay directly to the mercury ground state I assume?


Are you perhaps forgetting something in this list?
correct, directly to the ground state. I'm assuming I'm forgetting the gamma ray emission?
 
  • #5
Steve4Physics said:
How many different levels are of interest? How many possible transitions are there between these levels?
Can we just treat these as plain old photon emission transitions? i.e. ##E_{\gamma_1}=1.008 \, MeV - 0.412 \, MeV, \, E_{\gamma_2}=1.008 \, MeV - 0, \, E_{\gamma_3}=0.412 \, MeV - 0.##? If so than this question 1. is a lot easier than expected but also 2. leaves me even more confused as to what happens to those dang electrons and antineutrinos produced from ##\beta## decay!

Regardless, thanks :)
 
  • #6
oleosquarewave said:
correct, directly to the ground state. I'm assuming I'm forgetting the gamma ray emission?
If it decayed directly to the ground state, then there would be no gamma emission.
 
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  • #7
Steve4Physics said:
Hopefully you have a typo' and mean ##0.412MeV##.
I did whoops thanks for catching that
 
  • #8
oleosquarewave said:
leaves me even more confused as to what happens to those dang electrons and antineutrinos produced from β decay!
What about them? They are still produced in the initial decay.
 
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  • #9
Orodruin said:
If it decayed directly to the ground state, then there would be no gamma emission.
So it's decay to the excited state is what allows for the emission of gamma rays (once relaxation occurs), at least that's what I gather. Like you said, the ##Q## value isn't really of use here, but it is higher than the second excited state of ##^{198}\text{Hg}##. What happens to this extra energy? Kind of an extra-curricular question so no worries if there's not really a sufficient answer.
 
  • #10
The Q value being higher than the excitation energy means you have enough energy to actually excite that state. Whatever is not used to do that will end up as kinetic energy in the prompt decay products.
 
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  • #11
Orodruin said:
What about them? They are still produced in the initial decay.
The electron and antineutrino take energy to create, presumably taking away from the energy available for photon emission and constraining the energy of any emitted gamma rays, right? Or is the energy required to create them not at all related to the excitation energy levels of the nucleus? I'm assuming that this kind of question is a bit over my head, seeing as the textbook just gives us the excitation energies without telling us how they were found.
 
  • #12
oleosquarewave said:
The electron and antineutrino take energy to create, presumably taking away from the energy available for photon emission and constraining the energy of any emitted gamma rays, right?

Not right. They were created already in the prompt beta decay. They take exactly the energy they can while leaving the nucleus in whatever state it ends up in. What happens to the nucleus later is irrelevant to them.

oleosquarewave said:
Or is the energy required to create them not at all related to the excitation energy levels of the nucleus?
The energy required to create them is accounted for in the Q value.

oleosquarewave said:
I'm assuming that this kind of question is a bit over my head, seeing as the textbook just gives us the excitation energies without telling us how they were found.
 
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  • #13
Orodruin said:
Not right. They were created already in the prompt beta decay. They take exactly the energy they can while leaving the nucleus in whatever state it ends up in. What happens to the nucleus later is irrelevant to them.


The energy required to create them is accounted for in the Q value.
Thank you for the clarifications, much appreciated
 

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