A Brief Tutorial About Writing Nuclear Symbols

Fusion occurs between two light nuclides, resulting in a heavier nuclide plus (often, but not always) something else.

There is always an amount of energy given off during a fusion reaction. Examples #8-10 lay out the calculations involved in determining how much.

In many examples, the light nuclides that fuse are isotopes of hydrogen. I'll start there.

**Example #1:** Two deuterium nuclides fuse. What is the sole product?

**Solution:**

1) Write the reaction without the product being identified:

${\text{}}_{1}^{2}\text{H}$ + ${\text{}}_{1}^{2}\text{H}$ ---> _____

2) Atomic numbers must add up to be equal on both sides of the arrow. Mass numbers must add up to be equal on both sides of the arrow.

${\text{}}_{1}^{2}\text{H}$ + ${\text{}}_{1}^{2}\text{H}$ ---> ${\text{}}_{2}^{4}\text{He}$

3) By the way, energy is always given off in fusion reactions. A common symbol for the energy is Q:

${\text{}}_{1}^{2}\text{H}$ + ${\text{}}_{1}^{2}\text{H}$ ---> ${\text{}}_{2}^{4}\text{He}$ + Q

**Example #2:** Two deuterium nuclides fuse to form He-3. Write the full fusion equation.

**Solution:**

1) Write the reactants:

${\text{}}_{1}^{2}\text{H}$ + ${\text{}}_{1}^{2}\text{H}$ ---> _____ + _____ <--- Yup. Two products.

2) Now, include the He-3 that gets formed:

${\text{}}_{1}^{2}\text{H}$ + ${\text{}}_{1}^{2}\text{H}$ ---> ${\text{}}_{2}^{3}\text{He}$ + _____

3) We now need something with a mass number of one and an atomic number of zero. A neutron fits the bill exactly.

${\text{}}_{1}^{2}\text{H}$ + ${\text{}}_{1}^{2}\text{H}$ ---> ${\text{}}_{2}^{3}\text{He}$ + ${\text{}}_{0}^{1}\text{n}$

**Example #3:** Helium-4 is the major product when deuterium and tritium fuse. Write the full equation for this nuclear reaction.

**Solution:**

1) I'll start with just the reactants:

${\text{}}_{1}^{2}\text{H}$ + ${\text{}}_{1}^{3}\text{H}$ ---> _____

2) Let's put in just the He-4:

${\text{}}_{1}^{2}\text{H}$ + ${\text{}}_{1}^{3}\text{H}$ ---> ${\text{}}_{2}^{4}\text{He}$ + _____

3) A total mass number of 5 on the left must be reproduced on the right. (I added an energy term even though it wasn't called for in the problem statement:

${\text{}}_{1}^{2}\text{H}$ + ${\text{}}_{1}^{3}\text{H}$ ---> ${\text{}}_{2}^{4}\text{He}$ + ${\text{}}_{0}^{1}\text{n}$ + QNote that the total atomic number on each side were equal. Needing a zero for atomic number and a one for mass number can only be satisfied by a neutron.

**Example #4:** Describe, using a nuclear equation, the fusion of lithium-6 with hydrogen-2 to produce helium-4 and an alpha particle.

**Solution:**

1) The full equation:

${\text{}}_{3}^{6}\text{Li}$ + ${\text{}}_{1}^{2}\text{H}$ ---> 2 ${\text{}}_{2}^{4}\text{He}$Note: helium-4 and an alpha particle are the same thing.

2) Sometimes, you see the answer this way:

${\text{}}_{3}^{6}\text{Li}$ + ${\text{}}_{1}^{2}\text{H}$ ---> ${\text{}}_{2}^{4}\text{He}$ + ${\text{}}_{2}^{4}\text{He}$

3) Here is a slightly different fusion with lithium and hydrogen:

${\text{}}_{3}^{7}\text{Li}$ + ${\text{}}_{1}^{1}\text{H}$ ---> ${\text{}}_{2}^{4}\text{He}$ + ${\text{}}_{2}^{4}\text{He}$

**Example #5:** Wherein I simply write two fusion equations:

${\text{}}_{1}^{2}\text{H}$ + ${\text{}}_{2}^{3}\text{He}$ ---> ${\text{}}_{2}^{4}\text{He}$ + ${\text{}}_{1}^{1}\text{H}$${\text{}}_{2}^{3}\text{He}$ + ${\text{}}_{1}^{1}\text{H}$ ---> ${\text{}}_{2}^{4}\text{He}$ + ${\text{}}_{1}^{0}\text{e}$ <--- that's a positron

**Example #6:** Determine the missing nuclide:

${\text{}}_{1}^{2}\text{D}$ + ${\text{}}_{1}^{2}\text{D}$ ---> ${\text{}}_{1}^{3}\text{T}$ + _____

**Solution:**

1) First a comment:

Note how I used D and T rather than H. D means deuterium and T means tritium. Using H, deuterium is ${\text{}}_{1}^{2}\text{H}$ and tritium is ${\text{}}_{1}^{3}\text{H}$

2) (a) A mass number of 1 is needed to balance the total mass number on each side of the arrow. (b) An atomic number of 1 is needed to balance the total atomic number on each side of the arrow.

3) The result is:

${\text{}}_{1}^{2}\text{D}$ + ${\text{}}_{1}^{2}\text{D}$ ---> ${\text{}}_{1}^{3}\text{T}$ + ${\text{}}_{1}^{1}\text{H}$

**Example #7:** Write the isotope needed, by identifying A, Z, and X, to balance the following nuclear fusion reaction:

${\text{}}_{1}^{2}\text{H}$ + ${\text{}}_{1}^{2}\text{H}$ ---> ${\text{}}_{1}^{1}\text{H}$ + ${\text{}}_{\mathrm{Z}}^{\mathrm{A}}\text{X}$

**Solution:**

${\text{}}_{1}^{2}\text{H}$ + ${\text{}}_{1}^{2}\text{H}$ ---> ${\text{}}_{1}^{1}\text{H}$ + ${\text{}}_{1}^{3}\text{H}$

**Example #8:** Calculate the energy (in both J and MeV) released in the fusion reaction between deuterium and tritium. The reaction produces helium-4 and a neutron.

**Solution:**

1) The first thing to do is write the reaction (used also in example #3 above):

${\text{}}_{1}^{2}\text{H}$ + ${\text{}}_{1}^{3}\text{H}$ ---> ${\text{}}_{2}^{4}\text{He}$ + ${\text{}}_{0}^{1}\text{n}$ + Q

2) Secondly, we need the atomic masses of each nuclide:

Deuterium 2.014101778 u Tritium 3.016049282 u Helium-4 4.002603254 u neutron 1.008664916 u

3) Add the two reactants and the two products, then compute the difference:

(2.014101778 + 3.016049282) − (4.002603254 + 1.008664916)5.03015106 − 5.01126817

0.01888289 u <--- this is known as the mass defect

3) Next, we need to convert the mass defect (in atomic mass units) to kilograms. The conversion factor is 1 u = 1.6605402 x 10¯^{27} kg.

(0.01888289 u) (1.6605402 x 10¯^{27}kg/u) = 3.1355798 x 10¯^{29}kg

4) Use Einstein's famous formula to calculate the energy:

E = mc^{2}E = (3.1355798 x 10¯

^{29}kg) (299792458 m/s)^{2}E = 2.8181186 x 10¯

^{12}J

5) To determine the answer in MeV (megaelectron volt), we use the conversion factor of 1 J = 6.24150648 x 10^{12} MeV.

(2.8181186 x 10¯^{12}J) (6.24150648 x 10^{12}MeV / J) = 17.589 MeV

6) This reaction is also discussed in a YouTube video. He uses kJ/mol for the energy amount. My value of 17.589 MeV is for the reaction of one nuclide of H-2 with one nuclide of H-3 whereas his is for a mole of nuclides. The author does a fission example as well.

**Example #9:** Given the following fusion reaction (mentioned also in example #5):

${\text{}}_{1}^{2}\text{H}$ + ${\text{}}_{2}^{3}\text{He}$ ---> ${\text{}}_{2}^{4}\text{He}$ + ${\text{}}_{1}^{1}\text{H}$ + Q

Determine the energy (in MeV) released.

**Solution:**

1) Get the atomic masses:

Deuterium 2.014101778 u Helium-3 3.016029323 u Helium-4 4.002603254 u proton 1.007276467 u

2) Calculate the mass defect:

(2.014101778 + 3.016029323) − (4.002603254 + 1.007276467) = 0.02025138 u

3) Start with Einstein's formula:

E = mc^{2}E = (0.02025138 u) (c

^{2})E = (0.02025138 u-c

^{2}) (931.5 MeV / u-c^{2})See here for more info on the 931.5 MeV / u-c

^{2}value.E = 18.86 MeV

**Example #10:** Given the following fusion reaction (also used up above in example #2):

${\text{}}_{1}^{2}\text{H}$ + ${\text{}}_{1}^{2}\text{H}$ ---> ${\text{}}_{2}^{3}\text{He}$ + ${\text{}}_{0}^{1}\text{n}$ + Q

Determine the energy (in MeV) released.

**Solution:**

1) Mass of the reactants:

2.014101778 u + 2.014101778 u = 4.028203556 u

2) Mass of the products:

3.016029323 u + 1.008664916 u = 4.024694239 u

3) Determine the mass defect:

4.028203556 u − 4.024694239 u = 0.003509317 u

4) Convert u (mass) to MeV (energy):

E = mc^{2}Recall that 1 u = 931.5 MeV/c

^{2}(NOT just MeV, include the c^{2}. Remember that 931.5 MeV/c^{2}is a mass.)E = [(0.003509317 u) (931.5 MeV/c

^{2})] c^{2}E = 3.269 MeV