How to calculate gamma ray energies following beta decay?

In summary, to calculate gamma ray energies following beta decay, one must first identify the parent and daughter nuclei involved in the decay process. The energy difference between these two states, known as the Q-value of the decay, must be determined. This can be calculated using the mass-energy equivalence principle, where the masses of the parent and daughter nuclei, along with the emitted beta particle (electron or positron), are used to find the energy released during the decay. Once the Q-value is established, the gamma ray energies can be determined by considering the nuclear transitions that occur as the daughter nucleus moves to its ground state, with the energy of the emitted gamma rays corresponding to the differences in energy levels of the nuclear states involved.
  • #1
oleosquarewave
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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
 

FAQ: How to calculate gamma ray energies following beta decay?

What is beta decay?

Beta decay is a type of radioactive decay in which an unstable nucleus transforms into a more stable one by emitting a beta particle, which can be an electron (beta-minus decay) or a positron (beta-plus decay). This process often results in the emission of gamma rays as the nucleus transitions to a lower energy state.

How do gamma rays relate to beta decay?

During beta decay, the nucleus may transition to an excited state before settling into a more stable state. As it does so, it can release excess energy in the form of gamma rays. These gamma rays carry away energy that is not needed for the stability of the new nucleus, allowing it to reach its ground state.

What is the formula to calculate gamma ray energy?

The energy of the emitted gamma rays can be calculated using the equation: E = mc², where E is the energy, m is the mass defect (the difference in mass between the parent and daughter nuclei), and c is the speed of light. Alternatively, the energy can also be calculated by measuring the difference in energy levels of the nuclear states involved in the decay.

How do I find the mass defect for beta decay?

The mass defect can be found by determining the mass of the parent nucleus and the mass of the daughter nucleus along with the emitted beta particle. The mass defect is calculated as: Δm = m(parent) - [m(daughter) + m(beta particle)]. This mass defect can then be converted to energy using E = Δm * c².

What tools or methods can I use to measure gamma ray energies?

Gamma ray energies can be measured using various detectors, such as scintillation counters, semiconductor detectors, or Geiger-Müller tubes. Additionally, gamma spectroscopy allows for precise measurement of energy levels by analyzing the energy spectrum of emitted gamma rays, providing insights into the decay processes and energy transitions occurring in beta decay.

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