### Writing Nuclear ReactionsTen Examples

These ten examples are (mostly) of elements that have a single-digit atomic number. The five-example file uses elements from the heavier end of the periodic table, while the fifteen-example file uses alternates heavier elements with elements from (more-or-less) the middle of the periodic table.

As always, remember:

(a) atomic number is conserved, which means the total atomic number on each side is the same.
(b) mass number is conserved, which means the total mass number on each side is the same.

I also have several long/short form comments scattered through the examples.

Example #1: Beryllium-9 absorbs an alpha particle, producing a nucleus and a neutron. What is the nucleus?

Solution:

1) Write what we know:

${\text{}}_{4}^{9}\text{Be}$   +   ${\text{}}_{2}^{4}\text{He}$   --->   ____   +   ${\text{}}_{0}^{1}\text{n}$

2) Determine atomic number:

4 + 2 = x + 0

x = 6

3) Determine the mass number:

9 + 4 = x + 1

x = 12

4) Write the full equation:

${\text{}}_{4}^{9}\text{Be}$   +   ${\text{}}_{2}^{4}\text{He}$   --->     +   ${\text{}}_{0}^{1}\text{n}$

5) Write the short form:

${\text{}}_{4}^{9}\text{Be}$ (α, n)

Example #2: Be-8 absorbs an alpha particle, producing a nucleus plus gamma radiation. Write the full equation.

Solution:

1) Write the reactants:

${\text{}}_{4}^{8}\text{Be}$   +   ${\text{}}_{2}^{4}\text{He}$   --->

2) Atomic number and mass number:

4 + 2 = 6 (atomic no.)
8 + 4 = 12 (mass no.)

3) Write the full equation:

${\text{}}_{4}^{8}\text{Be}$   +   ${\text{}}_{2}^{4}\text{He}$   --->     +   γ

When there is seemingly one product, a gamma is often included as a second product. Sometimes, in the long form, the gamma is not written. However, it is almost always included in the short form.

4) Short form:

${\text{}}_{4}^{8}\text{Be}$ (α, γ)

Example #3: Cl-37 absorbs a proton to produce a nucleus. Write the equation.

Solution:

1) Write what we know:

${\text{}}_{17}^{37}\text{Cl}$   +   ${\text{}}_{1}^{1}\text{H}$   --->   _____

2) The product will have:

atomic number = 18 (from 17 + 1)
mass number of 38 (from 37 +1)

3) Write the full equation:

${\text{}}_{17}^{37}\text{Cl}$   +   ${\text{}}_{1}^{1}\text{H}$   --->   ${\text{}}_{18}^{38}\text{Ar}$   +   γ (Note the inclusion of gamma as the second product.)

4) Write the short form:

${\text{}}_{17}^{37}\text{Cl}$ (p, γ) ${\text{}}_{18}^{38}\text{Ar}$

Example #4: N-14 absorbs an alpha particle to produce a nucleus and protium. Write an equation.

Solution:

1) Write what is known:

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

Remember: protium is another name for the hydrogen-1 isotope.

2) Atomic number and mass number:

7 + 2 = x + 1
x = 8 (atomic number)

14 + 4 = x + 1
x = 17 (mass number)

3) Write the full equation:

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

4) Write the short form:

(α, p)

Example #5: A deuterium nucleus is hit by a gamma and splits apart. Write an equation for this reaction.

Solution:

(1) Write what we know:

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

2) We know from the text of the problem that the deuterium splits. Since deuterium contains a proton and a neutron, the split seems obvious.

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

3) Short form:

${\text{}}_{1}^{2}\text{H}$ (γ, n) ${\text{}}_{1}^{1}\text{H}$

4) This short form is also possible:

${\text{}}_{1}^{2}\text{H}$ (γ, p) ${\text{}}_{0}^{1}\text{n}$

I think the first one given is preferred, based on the idea that one isotope of hydrogen is transformed into another isotope of hydrogen (rather than into a free neutron).

Notice that this reaction is a fission. Almost all fission reactions involve a heavy nucleus, but there are some at the lighter end. See Example #7a just below for another example.

Example #6: Li-6 is impacted by a proton, forming two stable isotopes of helium.

Solution:

1) The reactants:

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

2) If you do not know the two stable isotopes of helium (He-3 and He-4), it can easily be looked up. The full equation is:

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

3) The short form:

${\text{}}_{3}^{6}\text{Li}$ (n, ${\text{}}_{2}^{3}\text{He}$) ${\text{}}_{2}^{4}\text{He}$

4) The convention in writing short form is to have the heavier product go outside the parentheses. If that convention was to be ignored, the following short form could be written:

${\text{}}_{3}^{6}\text{Li}$ (n, α) ${\text{}}_{2}^{3}\text{He}$

Example #7a: Li-6 is impacted by a neutron, leading to these products: the most abundant stable isotope of helium and an unstable isotope of hydrogen.

Solution:

1) Write what we know about the reactants:

${\text{}}_{3}^{6}\text{Li}$   +   ${\text{}}_{0}^{1}\text{n}$   --->

2) Include the most abundant stable isotope of helium:

${\text{}}_{3}^{6}\text{Li}$   +   ${\text{}}_{0}^{1}\text{n}$   --->   ${\text{}}_{2}^{4}\text{He}$

3) Determine the atomic number and mass number of the unstable isotope of hydrogen:

(3 + 0) − 2 = 1 (atomic number)
(6 + 1) − 4 = 3 (mass number)

4) Write the full equation:

${\text{}}_{3}^{6}\text{Li}$   +   ${\text{}}_{0}^{1}\text{n}$   --->   ${\text{}}_{2}^{4}\text{He}$   +   ${\text{}}_{1}^{3}\text{H}$

5) Short form:

${\text{}}_{3}^{6}\text{Li}$ (n, t) ${\text{}}_{2}^{4}\text{He}$

Reminder: t stands for tritium, an isotope of hydrogen.

6) Like Example #5, this reaction can be considered to be a fission. This one, however, is more in the classic fission mode where a nuclei is hit by a neutron and splits into two pieces. Notice also that in the fission of a heavy nuclide, several neutrons are ejected. That is not happening in this case.

Example #7b: Li-6 is impacted by a neutron, leading to these products: the least abundant stable isotope of helium and an unstable isotope of hydrogen.

Solution:

1) The reactants:

${\text{}}_{3}^{6}\text{Li}$   +   ${\text{}}_{0}^{1}\text{n}$   --->

2) Include the least abundant stable isotope of helium:

${\text{}}_{3}^{6}\text{Li}$   +   ${\text{}}_{0}^{1}\text{n}$   --->   ${\text{}}_{2}^{3}\text{He}$

The link in Example #6, step 2 provides information on the abundance in nature of the stable isotopes of helium.

3) Determine the atomic number and mass number of the unstable isotope of hydrogen:

(3 + 0) − 2 = 1 (atomic number)
(6 + 1) − 3 = 4 (mass number)

4) Write the full equation:

${\text{}}_{3}^{6}\text{Li}$   +   ${\text{}}_{0}^{1}\text{n}$   --->   ${\text{}}_{2}^{3}\text{He}$   +   ${\text{}}_{1}^{4}\text{H}$

5) Short form:

${\text{}}_{3}^{6}\text{Li}$ (n, ${\text{}}_{2}^{3}\text{He}$) ${\text{}}_{1}^{4}\text{H}$

Example #8: Oxygen-16 absorbed a proton. An alpha particle is ejected and another nucleus remains. Write the full equation for this reaction.

Solution:

1) Write the given information:

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

Sometimes, when the word proton is used, the equation writer will use ${\text{}}_{1}^{1}\text{p}$ rather than ${\text{}}_{1}^{1}\text{H}$.

It's basically a stylistic choice.

2) Atomic number of the unknown nuclide:

8 + 1 = x + 2

x = 7

3) Mass number of the unknown nuclide:

16 + 1 = y + 4

y = 13

4) The full equation:

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

5) Short form:

(p, α)

Example #9: Al-26 with an alpha particle to produce P-30.

Solution:

1) Suppose you wrote this for the answer:

${\text{}}_{13}^{26}\text{Al}$   +   ${\text{}}_{2}^{4}\text{He}$   --->   ${\text{}}_{15}^{30}\text{P}$

You might get not get full credit for the answer, because it did not include a gamma on the right-hand side.

2) Here's the short form with the gamma included:

${\text{}}_{13}^{26}\text{Al}$ (α, γ) ${\text{}}_{15}^{30}\text{P}$

All nuclear reactions emit some radiation. However, the writing of a gamma as a product is usually limited to those nuclear reactions that produce only one nuclide on the right-hand side. This then allos a full short form to be written.

Example #10: Cu-63 absorbs a proton and produces Cl-38, a neutron and another nucleus.

Solution:

1) This is what the problem tells us:

${\text{}}_{29}^{63}\text{Cu}$   +   ${\text{}}_{1}^{1}\text{H}$   --->   ${\text{}}_{17}^{38}\text{Cl}$   +   _____   +   ${\text{}}_{0}^{1}\text{n}$

2) We see that 13 is required for the atomic number of the unknown nuclide and that 25 is the mass number. Write the complete equation:

${\text{}}_{29}^{63}\text{Cu}$   +   ${\text{}}_{1}^{1}\text{H}$   --->   ${\text{}}_{17}^{38}\text{Cl}$   +   ${\text{}}_{13}^{25}\text{Al}$   +   ${\text{}}_{0}^{1}\text{n}$

3) There is no short form. This reaction qualifies as a fission. There are some fission reactions that do not involve the heavier elements, it's just that there are not very many of them.

Another fission example would be Be-8 fissioning into two alpha particles. This is actually an important step in the generation of energy in a star.

Also, fission of lighter elements is 100% 'induced fission.' Heavy elements are subject to 'induced fission' as well, but they also undergo 'spontaneous fission.