ChemTeam: Writing Nuclear Reactions: Fifteen Examples Equations

Writing Nuclear Reactions
Fifteen Examples

Writing Nuclear Equations: Five Examples      Writing Fission Equations
Writing Nuclear Equations: Ten Examples      Writing Fusion Equations
Converting between long and short forms of nuclear reactions      Radioactivity Menu

A Brief Tutorial About Writing Nuclear Symbols

These fifteen examples alternate between the heavier elements and elements taken (more-or-less) from the middle of the periodic table. The five-element file uses elements from the heavy end of the periodic table while the ten-example file uses mostly light elements.

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: U-238 absorbs a neutron to produce a nucleus and a beta particle.

Solution:

1) Write what we know from the text of the problem:

$_{92}^{238} U$ + $_{0}^{1} n$ ---> _____ + $_{−1}^{0} e$

2) Determine atomic number and mass number of the unknown nuclide:

at. number ---> 93 must be the atomic number of the unknown nuclide
mass number ---> 239 must be the mass number of the unknown nuclide
Remember the word 'conserved.'

3) Include the nuclide in the reaction equation:

$_{92}^{238} U$ + $_{0}^{1} n$ ---> $_{93}^{239} Np$ + $_{−1}^{0} e$

4) Here is the short form (sometimes the beta is shown as β¯):

$_{92}^{238} U$ (n, β) $_{93}^{239} Np$

5) By the way, Np-239 decays with a half-life a bit less than 2.5 days. It decays by beta:

$_{93}^{239} Np$ ---> $_{94}^{239} Pu$ + $_{−1}^{0} e$
Pu-239 was used in the atomic bomb exploded over Nagasaki and is also used in the fission trigger of a thermonuclear bomb.

Example #2: Al-27 with deuterium to produce an alpha particle and another nucleus.

Solution:

1) Write what we know:

$_{13}^{27} Al$ + $_{1}^{2} H$ ---> _____ + $_{2}^{4} He$

2) Determine atomic and mass number for the unknown nuclide:

13 + 1 = x + 2 ---> x = 12 (atomic number)
27 + 2 = y + 4 ---> y = 25 (mass number)

3) Write the full equation:

$_{13}^{27} Al$ + $_{1}^{2} H$ ---> $_{12}^{25} Mg$ + $_{2}^{4} He$

4) Write the short form:

$_{13}^{27} Al$ (d, α) $_{12}^{25} Mg$

Example #3: U-238 absorbs a proton to produce a nucleus and gamma radiation.

Solution:

1) Write what we know from the text of the problem:

$_{92}^{238} U$ + $_{1}^{1} p$ ---> _____ + γ

2) Determine atomic number and mass number of the unknown nuclide:

92 + 1 give 93 for the atomic number
238 + 1 gives 239 for the mass number of the unknown nuclide
Remember, gamma counts for zero in atomic number and mass number.

3) Include the nuclide in the reaction equation:

$_{92}^{238} U$ + $_{1}^{1} p$ ---> $_{93}^{239} U$ + γ
Often, you will see $_{1}^{1} H$ used for the proton.
Since both are used, you have to know both.

4) Here is the short form:

$_{92}^{238} U$ (p, γ) $_{93}^{239} Np$

Example #4: Cu-63 with a nucleus producing Zn-63 and a neutron.

Solution:

1) Write what the problem statement tells us:

$_{29}^{63} Cu$ + _____ ---> $_{30}^{63} Zn$ + $_{0}^{1} n$

2) The atomic number and the mass number of the unknown:

The atomic number must be one, in order to satisfy the 64 present on the right-hand side.
The mass number must also be 1, in order to get a total of 30 on each side.

3) The full equation:

$_{29}^{63} Cu$ + $_{1}^{1} H$ ---> $_{30}^{63} Zn$ + $_{0}^{1} n$

4) Short form:

$_{29}^{63} Cu$ (p, n) $_{30}^{63} Zn$

Example #5: Pu-239 can be produced by bombarding U-238 with alpha particles. Write the nuclear equation for this reaction. Also, identify how many neutrons are released.

Solution:

1) I propose to write just the left-hand side:

$_{92}^{238} U$ + $_{2}^{4} He$ --->
Even though the plural is used in the problem statement ("alpha particles"), only one alpha particle is absorbed into a uranium-238 nucleus.

2) Add in the right-hand side:

$_{92}^{238} U$ + $_{2}^{4} He$ ---> $_{94}^{239} Pu$ + _____

3) Determine the number of neutrons produced:

(238 + 4) − 239 = 3
I can ignore the atomic number because I know only neutrons (at. num. = 0) are involved.

4) Write the full equation:

$_{92}^{238} U$ + $_{2}^{4} He$ ---> $_{94}^{239} Pu$ + 3 $_{0}^{1} n$

5) Short form:

$_{92}^{238} U$ (α, 3n) $_{94}^{239} Pu$

Example #6: Bombardment of Al-27 with an alpha particle produces P-30 and another particle

Solution:

1) Write what the problem gives us:

$_{13}^{27} Al$ + $_{2}^{4} He$ ---> $_{15}^{30} P$ + _____
By the way, sometimes $_{2}^{4} α$ is used in the equation. Sometimes, only the letter alpha (α) is used.
The ChemTeam prefers to use $_{2}^{4} He$ in the long form and α in the short form.

2) Atomic number and mass number for the missing nuclide:

Zero for the atomic number seems pretty obvious. For the mass number, 27 plus 4 and then minus 30 gives 1.

3) Include the neutron in the answer:

$_{13}^{27} Al$ + $_{2}^{4} He$ ---> $_{15}^{30} P$ + $_{0}^{1} n$

4) Short form:

$_{13}^{27} Al$ (α, n) $_{15}^{30} P$

Example #7: Describe, using a nuclear equation, U-238 absorbing a neutron to form Pu-239 accompanied by the emission of a beta particle.

Solution:

1) Write just the left-hand (reactant) side:

$_{92}^{238} U$ + $_{0}^{1} n$ --->

2) We know that Pu-239 is a product. Include it:

$_{92}^{238} U$ + $_{0}^{1} n$ ---> $_{94}^{239} Pu$

3) Examine the atomic number situation:

The 94 on the right-hand side must be offset by 2 negatives to get the 92 on the left-hand side.
That means two beta particles are required.

4) Write the full equation:

$_{92}^{238} U$ + $_{0}^{1} n$ ---> $_{94}^{239} Pu$ + 2 $_{−1}^{0} e$
Notice how the wording of the problem is misleading. The wording is singular, but two beta particles are required.

5) Short form:

$_{92}^{238} U$ (n, 2β¯) $_{94}^{239} Pu$

Example #8: Given the following short form:

$_{27}^{59} Co$ (___, γ) $_{27}^{60} Co$

Determine the missing nuclide.

Solution:

1) The atomic number does not change. That means the missing nuclide has zero for the atomic number.

2) The mass number goes up by 1. Coupled with the zero change in the atomic number, the missing nuclide is a neutron.

3) The complete short form:

$_{27}^{59} Co$ (n, γ) $_{27}^{60} Co$

4) The long form:

$_{27}^{59} Co$ + $_{0}^{1} n$ ---> $_{27}^{60} Co$ + γ

Example #9: U-238 is bombarded by alpha particles, resulting in a new nucleus being formed and two neutrons being ejected. Write the long form as well as the short form for this reaction.

Solution:

I'll just list the answers.

$_{92}^{238} U$ + $_{2}^{4} He$ ---> $_{94}^{240} Pu$ + 2 $_{0}^{1} n$
$_{92}^{238} U$ (α, 2n) $_{94}^{240} Pu$

Example #10: Fill in the blank.

$_{48}^{113} Cd$ (___, γ) $_{48}^{113} Cd$

Solution:

1) No change in atomic number. No change in mass number.

2) Only one thing does not change atomic/mass number.

3) Write the complete short form:

$_{48}^{113} Cd$ (γ, γ) $_{48}^{113} Cd$

4) By the way, a neutrino does not change the atomic/mass number either. However, neutrinos are only discussed in nuclear decay reactions such as beta or positron decay. Neutrinos do not appear in the type of reactions being discussed here.

5) I won't bother with the long form.

Example #11: In which I simply write a full nuclear equation and its short form.

$_{95}^{240} Am$ + $_{2}^{4} He$ ---> $_{97}^{243} Bk$ + $_{0}^{1} n$
$_{95}^{240} Am$ (α, n) $_{97}^{243} Bk$
By the way, Am-241 is used in household smoke detectors.

Example #12: U-238 is bombarded with C-12 to produce a nuclide and 4 neutrons.

Solution:

1) Write what we know:

$_{92}^{238} U$ + $_{6}^{12} C$ ---> _____ + 4 $_{0}^{1} n$

2) Determine atomic number:

92 + 6 = 98
The nuclide is an isotope of Californium (Yay! The ChemTeam's home state.)

3) Determine the mass number:

(238 + 12) − (x + 4)
x = 246

4) Write the full equation:

$_{92}^{238} U$ + $_{6}^{12} C$ ---> $_{98}^{246} Cf$ + 4 $_{0}^{1} n$

5) The short form:

$_{92}^{238} U$ ( $_{6}^{12} C$ , 4n) $_{98}^{246} Cf$

Example #13: U-238 is bombarded with C-12 to produce a nuclide and 6 neutrons.

Solution:

1) The full equation:

$_{92}^{238} U$ + $_{6}^{12} C$ ---> $_{98}^{244} Cf$ + 6 $_{0}^{1} n$

2) The short form:

$_{92}^{238} U$ ( $_{6}^{12} C$ , 6n) $_{98}^{244} Cf$

Example #14: Pu-239 is bombarded by alpha articles. A neutron and what else is produced? Use a method of isotopic notation other than the standard one.

Solution:

1) Using the problem statement, write what we know:

94-Pu-239 + 2-He-4 ---> _____ + 0-n-1

2) The atomic number of the unknown nuclide must be 96 and its mass number must be 242.

3) Write the complete equation:

94-Pu-239 + 2-He-4 ---> 96-Cm-242 + 0-n-1

4) Write the short form:

94-Pu-239 (α, n) 96-Cm-242
I have never, ever seen a short form written as I just did. All the examples I have ever seen use standard isotopic notation. As follows:
$_{94}^{239} Pu$ (α, n) $_{96}^{242} Cm$

Example #15: Cm-242 is bombarded with alpha particles, producing a nucleus and a neutron. What is the nucleus?

Solution:

1) From the problem statement:

$_{96}^{242} Cm$ + $_{2}^{4} He$ ---> _____ + $_{0}^{1} n$

2) Atomic number and mass number of the unknown nuclide:

Z must be 98
A must be 245

3) The long form:

$_{96}^{242} Cm$ + $_{2}^{4} He$ ---> $_{98}^{245} Cf$ + $_{0}^{1} n$

4) The short form:

$_{96}^{242} Cm$ (α, n) $_{98}^{245} Cf$

Writing Nuclear Equations: Five Examples	Writing Fission Equations
Writing Nuclear Equations: Ten Examples	Writing Fusion Equations
Converting between long and short forms of nuclear reactions	Radioactivity Menu

Writing Nuclear ReactionsFifteen Examples

Writing Nuclear Reactions
Fifteen Examples