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. 𝐾 → 𝐴𝑟 + 𝑒! + 𝜈!!"

!"!"!" 𝑄 = 𝑀!𝑐! − (𝑀!𝑐! +𝑚!𝑐!)−𝑚!𝑐! = 𝑀!𝑐! −𝑀!𝑐! − 2𝑚!𝑐!

𝑄 = 39.964000− 39.962384 𝑢 ∗931.5𝑀𝑒𝑉

𝑢 − 2 ∗ 0.5109989 = 0.4833𝑀𝑒𝑉 𝐾 → 𝐴𝑟 +!"

!"!"!" 𝜈! 𝑄 = 𝑀!𝑐! −𝑀!𝑐!

𝑄 = 39.964000− 39.962384 𝑢 ∗931.5𝑀𝑒𝑉

𝑢 = 1.505𝑀𝑒𝑉 𝐾 → 𝐶𝑎 + 𝑒! + 𝜈!!"

!"!"!" 𝑄 = 𝑀!𝑐! − (𝑀!𝑐! −𝑚!𝑐!)−𝑚!𝑐! = 𝑀!𝑐! −𝑀!𝑐!

𝑄 = 39.964000− 39.962581 𝑢 ∗931.5𝑀𝑒𝑉

𝑢 = 1.322𝑀𝑒𝑉 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 proton-21 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 𝑆𝑒 → 𝐵𝑟 + 𝑒! + 𝜈!!"

!"!"!" 𝑄 = 𝑀!𝑐! − (𝑀!𝑐! −𝑚!𝑐!)−𝑚!𝑐! = 𝑀!𝑐! −𝑀!𝑐!

𝑄 = 81.916697− 81.916802 𝑢 ∗931.5𝑀𝑒𝑉

𝑢 < 0𝑀𝑒𝑉 This would violate conservation of energy 𝑆𝑒 → 𝐾𝑟 + 2𝑒! + 2𝜈!!"

!"!"!" 𝑄 = 𝑀!𝑐! − (𝑀!𝑐! − 2𝑚!𝑐!)− 2𝑚!𝑐! = 𝑀!𝑐! −𝑀!𝑐!

𝑄 = 81.916697− 81.913481 𝑢 ∗931.5𝑀𝑒𝑉

𝑢 > 0𝑀𝑒𝑉 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.