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Opus in profectus

Nucleosynthesis

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Practice

practice problem 1

Hydrogen fusion in the Sun is a multistep reaction, but the net result is that four hydrogen atoms fuse into one helium atom (plus a bunch of junk).

411H → 42He + 2(0+1e + 00γ + 00ν)

The mass of the Sun is 1.99 × 1030 kg, 91% of which is hydrogen. Its power output is 3.85 × 1026 W. Determine…

  1. the mass of four hydrogen atoms
  2. the mass defect when four hydrogen atoms fuse into one helium atom (in atomic mass units and megaelectronvolts)
  3. the rate at which the Sun's mass is decreasing
  4. the total mass destroyed if all the Sun's hydrogen were converted into helium
  5. the expected lifetime of the Sun (assuming its power output will remain constant)

solution

  1. The mass of four hydrogen atoms is…

    4(1.007825 u) = 4.031300 u

  2. The mass difference between four hydrogen atoms and one helium atom is…

    4(1.007825 u) − (4.00260 u) = 0.02870 u

    (0.02870 u)(931 MeV/u) = 26.72 MeV

  3. The rate at which mass is destroyed is related to the rate at which energy is produced.

    m  =  P
    t c2
     ⇐ 
    P =  E  =  mc2
    t t
    m  =  3.85 × 1026 W
    t (3.00 × 108 m/s)2
    m  =  4.29 × 109 kg/s
    t

    About four million tons of the Sun vanishes every second.

  4. The fraction of the Sun's mass lost if every hydrogen nucleus was used to produce helium is the same as the the ratio of the mass defect to the original mass.

    destroyed mass  =  m  =  0.0287 u
    initial mass (0.91)(1.99 × 1030 kg) 4.0313 u

    m = 1.29 × 1028 kg

  5. Setting up another proportion finishes the problem.

    m  =  4.29 × 109 kg  =  1.29 × 1028 kg
    t 1 s t

    t = 3.01 × 1018 s = 95 × 109 years

    This is a not a good estimate, however. Although extremely hot in human terms, most of the Sun is just too cool for hydrogen to fuse into helium. Only the central core is hot enough and dense enough. Thus, it is estimated that not more than ten percent of the Sun's total hydrogen will ever be available for thermonuclear fusion. Ten percent of 95 billion years 9.5 billion years, which is still a good long time. The Earth is some 4.5 billion years old already, placing us somewhere in the middle of the Sun's life. The sun is a middle-aged star.

practice problem 2

Let's turn lead into gold. Read this.

The dreams of medieval alchemists have nearly come true in the modern era of accelerators and nuclear reactions. For example, one can fuse two metallic atoms, a germanium‑74 (7432Ge) projectile with a tin‑124 (12450Sn) target. When the 7432Ge has a kinetic energy of 300 MeV, which corresponds to about 9% of the speed of light, a 19882Pb* nucleus is created in the nuclear fusion.

That initial 19882Pb* nucleus has a mass number equal to the sum of the Ge projectile's and the Sn target's mass numbers. The asterisk indicates that the nucleus — which is created "hot" with an excitation energy of about 50 MeV, corresponding to a temperature of more than 1010 K — is a compound nucleus, one that is not fully bound. It promptly evaporates several neutrons to cool down, and different Pb isotopes, called fusion-evaporation residues, are created in the process.

In the 19882Pb* example, the evaporation of four neutrons to produce 19482Pb accounts for about 60% of the evaporation residues. Within a few tens of minutes, beta decay transforms the 19482Pb into thallium‑194 (19481Tl). Then a second beta decay turns 19481Tl into mercury‑194 (19480Hg), a nucleus with a 520 year half life. The nuclear alchemist seeking Au has to wait quite some time for the next beta decay into 19479Au. Unfortunately, 19479Au is an unstable isotope, with a half life of only 38 hours.

The beta decay of 19479Au does create stable and even more precious platinum‑194 (19478Pt), but at typical beam intensities used in current SHE experiments, continuous irradiation of 12450Sn with 7432Ge would produce only 1 g of stable 19478Pt in about 100 million years. So nuclear alchemy is possible in principle, but it is definitely not a wise capital venture. However, experiments creating superheavy nuclei are much more rewarding in terms of scientific gain.

Oganessian and Rykaczewski, 2015

Now, write out the reactions in symbolic form.

solution

And here we go. Note that the only way this works out is if the beta decays are all positive beta decays (a.k.a. positron emission). The authors did not state or imply this in their description.

7432Ge + 12450Sn → 19882Pb*

19882Pb* → 19482Pb + 410n

19482Pb → 19481Tl + 0+1e

19481Tl → 19480Hg + 0+1e

19480Hg → 19479Au + 0+1e

19479Au → 19478Pt + 0+1e

The decay reactions, everything after the first line, could be written something like this.

19882Pb*    4͢n    19482Pb    β͢+    19481Tl    β͢+    19480Hg    β͢+    19479Au    β͢+    19478Pt

Yuck, that looks like crap in every browser I've tested it on. This style is fine for handwriting on paper or a blackboard, but it is not HTML friendly.

practice problem 3

Write something.

solution

Answer it.

practice problem 4

Write something.

solution

Answer it.