Physics 249 Homework 10
Due Dec 7th
1) a) Show that Potassium 40 can energetically undergo beta-, beta+ and electron capture.
b) Using the energy level diagrams from lecture 34 discuss why it is possible for this
isotope to undergo all three interactions.
40
19𝐾
+
→ 40
18𝐴𝑟 + 𝑒 + 𝜈𝑒
𝑄 = 𝑀𝑖 𝑐 2 − (𝑀𝑓 𝑐 2 + 𝑚𝑒 𝑐 2 ) − 𝑚𝑒 𝑐 2 = 𝑀𝑖 𝑐 2 − 𝑀𝑓 𝑐 2 − 2𝑚𝑒 𝑐 2
931.5𝑀𝑒𝑉
𝑄 = (39.964000 − 39.962384)𝑢 ∗ (
) − 2 ∗ 0.5109989 = 0.4833𝑀𝑒𝑉
𝑢
40
19𝐾
→ 40
18𝐴𝑟 + 𝜈𝑒
𝑄 = 𝑀𝑖 𝑐 2 − 𝑀𝑓 𝑐 2
931.5𝑀𝑒𝑉
𝑄 = (39.964000 − 39.962384)𝑢 ∗ (
) = 1.505𝑀𝑒𝑉
𝑢
40
19𝐾
−
→ 40
20𝐶𝑎 + 𝑒 + 𝜈̅𝑒
𝑄 = 𝑀𝑖 𝑐 2 − (𝑀𝑓 𝑐 2 − 𝑚𝑒 𝑐 2 ) − 𝑚𝑒 𝑐 2 = 𝑀𝑖 𝑐 2 − 𝑀𝑓 𝑐 2
931.5𝑀𝑒𝑉
𝑄 = (39.964000 − 39.962581)𝑢 ∗ (
) = 1.322𝑀𝑒𝑉
𝑢
b) This part has not been graded. It required you to incorporate arguments on the energy
level diagrams and the effect of proton repulsion with an even/odd proton or neutron
nuclei discussion presented, though not discussed in detail, in lecture 32.
Cases where both beta+ or electron capture and beta- decay are rare. In this case two
important effects are present.
First consider the proton and neutron configurations.
With 18,19, or 20 protons the highest level is 3d3/2. (note I am using the notation where
the first number indicates the shell)
With 21 neutrons (initial state) or 22 neutrons (after beta + decay) the highest level is
4f7/2 while with 20 neutrons ( beta - ) the highest level is 3d3/2 (though a higher energy
proton level is added).
For beta + decay and electron capture to occur the last proton in the level must be split
high to be very near the 4f7/2 neutron level.
With similar energy levels the even vs odd effect discussed in lecture 32 comes into
effect. Even configurations are known to be more stable. 40 K is an odd-odd 19 proton21 neutron configuration. Both decay channels lead to even-even configurations, which
are more stable. This is due to an effect known as pairing where it is more energetically
favorable to have pairs of neutrons or protons due to a property of the nuclear strong
force. For an example of unpaired binding look at the energy levels of the deuteron
presented in lecture 33, which has particularly small binding energy compared to typical
nuclear binding energies. This demonstrates that pairing can be a significant effect.
The phenomena of simultaneously allowed beta + or electron capture and beta – decay is
restricted to odd-odd nuclei such as K 40.
2) Show why Selenium 82 decays via double beta decay rather than single beta decay.
Why is the lifetime so long for this decay?
Consider
82
82
−
34𝑆𝑒 → 35𝐵𝑟 + 𝑒 + 𝜈̅𝑒
𝑄 = 𝑀𝑖 𝑐 2 − (𝑀𝑓 𝑐 2 − 𝑚𝑒 𝑐 2 ) − 𝑚𝑒 𝑐 2 = 𝑀𝑖 𝑐 2 − 𝑀𝑓 𝑐 2
931.5𝑀𝑒𝑉
𝑄 = (81.916697 − 81.916802)𝑢 ∗ (
) < 0𝑀𝑒𝑉
𝑢
This would violate conservation of energy
82
34𝑆𝑒
−
→ 82
36𝐾𝑟 + 2𝑒 + 2𝜈̅𝑒
2
2
𝑄 = 𝑀𝑖 𝑐 − (𝑀𝑓 𝑐 − 2𝑚𝑒 𝑐 2 ) − 2𝑚𝑒 𝑐 2 = 𝑀𝑖 𝑐 2 − 𝑀𝑓 𝑐 2
931.5𝑀𝑒𝑉
𝑄 = (81.916697 − 81.913481)𝑢 ∗ (
) > 0𝑀𝑒𝑉
𝑢
This decay involves two simultaneous beta decays so you should square the probability
for the process to happen, which will substantially increase the lifetime.
3) 7 Be decays exclusively via electron capture. Comment on the change in the halflife
compared to the original atom if you remove one by one in sequence all of the electrons.
State and justify how the half-life changes in general terms without performing any
calculations.
Comparing diagrams on pages 291 and 292. The 1s n=1 l=0 orbit has a large probability
distribution at low radius while the 2s n=2 l=0 orbit has two probability lobes and the one
at low radius is order ¼ the probability per under radius (¼ the height).
Electron capture will occur for electrons that have significant probability to exist in the
nucleus where they can interact weakly with a proton.
For the two 2s electron has significantly less probability to exist there. Also removing
them will remove a small amount of radial shielding, also lower since they have low
probability to exist at low radius, and will provide a counter effect of increasing the Zeff
for the remaining electrons and increasing the probability they will interact. The overall
effect of removing the first two electrons will be small.
Removing the 3rd electron will remove ½ of the probability for interacting since there will
only be 1 electron left rather then two. However, it increase the effective Z by removing
some shielding. In fact according to page 306 the ionization energy is about twice as
large for the last electron in a 1S orbital compared to the second to last orbital indicating
that it is bound into an orbit of half size counteracting the reduction on probability due to
having one less electron that can undergo electron capture.
4) Compute the total energy released in the proton-proton fusion cycle. How much
energy would be released in Joules if you fused a mole of hydrogen gas. Compare this
the energy that is produced from burning the same mass of coal (coal makes 3.15x107
J/kg) .
The Q values for each stage are given on pages 551 and 552 though can simply be
calculated by subtracting the masses of the particles involved.
For the first step. The deuteron is very massive compared to the positron and neutrino
and will more at non-relativistic speeds. Using p=mv the velocity of the deuteron should
classically be order 4000 times less then the electron and it’s kinetic energy similarly
4000 times less. We will neglect this and assume on average the Q value energy is split
between the positron and the neutrino. The neutrino energy is lost and the ½ the energy is
with the positron. The interaction involves freeing one electron which will annihilate
with the positron contributing an additional 1.022 MeV of energy.
The energy at each step is
1: 1.23 MeV x 2 = 2.46 MeV
2: 5.49 MeV x 2 = 10.98 MeV
3: 12.86 MeV
Total: 26.3 MeV
The process leaves 2 H atoms over for the next process after many repetitions it takes 4 H
atoms per cycle or 2 H2 molecules or we get 13.15MeV per H2 molecule.
Given 6.022x10+23 H2 molecules in a mole you get: 7.92x10+24 MeV
1 mole of H2 gas will with 0.002 Kg and 0.002 Kg of coal will yield 63000J or
3.93x1017 MeV
a factor of 20 million.