Writing Nuclear Reactions
Five Examples

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

A Brief Tutorial About Writing Nuclear Symbols

These five examples are all drawn from heavier nuclides because I wanted to make some Glenn Seaborg comments. The ten-example file (mostly) uses light elements while the fifteen-example file alternates heavier elements with elements that are more-or-less in 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: U-238 is bombarded with another nuclide to produce Fm-249 and five neutrons

Solution:

1) Write as much of the reaction as possible:

${}_{\mathrm{ 92}}^{238}\text{U}$   +   _____   --->   ${}_{100}^{249}\text{Fm}$   +   5 ${}_0^1\text{n}$

2) Determine the atomic number:

100 − 92 = 8

The missing nuclide is an isotope of oxygen.

3) Determine the mass number:

(249 +5) = (238 + x)

x = 16

4) Write the full equation:

${}_{\mathrm{ 92}}^{238}\text{U}$   +   ${}_{\mathrm{ 8}}^{16}\text{O}$   --->   ${}_{100}^{249}\text{Fm}$   +   5 ${}_0^1\text{n}$

5) Write the short form:

${}_{\mathrm{ 92}}^{238}\text{U}$ (${}_{\mathrm{ 8}}^{16}\text{O}$, 5n) ${}_{100}^{249}\text{Fm}$

Example #2: Cf-249 is bombarded with an isotope of oxygen to produce Sg-263 and 4 neutrons. Write the full equation for this reaction.

Solution:

1) Write what we know:

${}_{\mathrm{ 98}}^{249}\text{Cf}$   +   ${}_{\mathrm{ 8}}^{\mathrm{??}}\text{O}$   --->   ${}_{106}^{263}\text{Sg}$   +   4 ${}_0^1\text{n}$

2) Determine the mass number:

249 + ?? = 263 + 1 + 1 + 1 + 1 (just emphasizing 4 neutrons)

?? = 18

3) Write the full equation:

${}_{\mathrm{ 98}}^{249}\text{Cf}$   +   ${}_{\mathrm{ 8}}^{18}\text{O}$   --->   ${}_{106}^{263}\text{Sg}$   +   4 ${}_0^1\text{n}$

4) The short form:

${}_{\mathrm{ 98}}^{249}\text{Cf}$ (${}_{\mathrm{ 8}}^{18}\text{O}$, 4n) ${}_{106}^{263}\text{Sg}$

5) Some comments:

(a) the equation in this question is the discovery equation for element 106. It was discovered by a team led by Albert Ghiorso, the discoverer (or co-discoverer) of 12 elements, ranging from element 95 to element 106.

(b) "Nuclear Reactions and Synthesis of New Transuranium Species" is a survey written by Glenn Seaborg in 1983. (When you get there, click the green-shaded link to the left side of the text. The one that says 'View Conference.') Seaborg led the work in tranuranium elements, discovering plutonium in 1940 but not publishing his work until 1948 (due to WWII-related secrecy). Seaborg shared the 1951 Nobel Prize in Chemistry for this work.

(c) The ChemTeam met Seaborg about 1984. In response to a question (not from the ChemTeam), Seaborg stated that his personal opinion was that his doctoral advisor, Gilbert Newton Lewis, had committed suicide and not due to a laboratory accident. By the way, Seaborg was the one to find the body.

4) Coincidently, the ChemTeam's father was the one to find the body of a colleague who had committed suicide in the lab by use of cyanide. An autopsy showed the man to have terminal cancer, a fact that the man had kept secret from his colleagues.

Example #3: A ${}_{\mathrm{}}^{250}\text{Cf}$ nucleus absorbs a ${}_{\mathrm{}}^{11}\text{B}$ nucleus and releases four neutrons. Identify the nuclide that remains. Write the short form first, followed by the long form.

Discussion:

Note the lack of an atomic number in the problem statement. This is because the atomic number is implied by the symbol of the element, making the explicit use of the atomic number redundant. However, the ChemTeam has decided to use the full isotopic notation.

Solution:

1) Identify the nuclide that remains:

atomic number ---> (98 + 5) − (4 * 0) = 103
mass number ---> (250 + 11) − 4 = 257

${}_{103}^{257}\text{Lr}$

2) Write the reactants and the products:

(a) the reactants are ${}_{\mathrm{ 98}}^{250}\text{Cf}$  and  ${}_{\mathrm{ 5}}^{11}\text{B}$

(b) the products are four neutrons and  ${}_{103}^{257}\text{Lr}$

3) The short form:

${}_{\mathrm{ 98}}^{250}\text{Cf}$ (${}_{\mathrm{ 5}}^{11}\text{B}$, 4n) ${}_{103}^{257}\text{Lr}$

4) The long form:

${}_{\mathrm{ 98}}^{250}\text{Cf}$   +   ${}_{\mathrm{ 5}}^{11}\text{B}$   --->   ${}_{103}^{257}\text{Lr}$   +   4 ${}_0^1\text{n}$

Example #4: In which I simply write the short form for a nuclear reaction:

${}_{\mathrm{ 99}}^{253}\text{Es}$ (α, n) ${}_{101}^{256}\text{Md}$

Example #5: The discovery reaction for element 109 involved bombarding a Bi-209 target with a Fe-58 projectile. A neutron was ejected. Write the long form and short form equations for this reaction.

Solution:

1) Long form:

${}_{\mathrm{ 83}}^{209}\text{Bi}$   +   ${}_{26}^{58}\text{Fe}$   --->   ${}_{109}^{266}\text{Mt}$   +   ${}_0^1\text{n}$

2) Short form:

${}_{\mathrm{ 83}}^{209}\text{Bi}$ (${}_{26}^{58}\text{Fe}$, n) ${}_{109}^{266}\text{Mt}$

3) Up above, in Example #2, I referenced a document by Glenn Seaborg. In it, he refers to element 109 on page seven as element 109 or as ²⁶⁶₁₀₉, never once referring to it with the IUPAC-approved temporary name of unnilennium, a name based on the Latin word for one hundred nine.

Bonus Example: Pu-239 can be used to synthesize Am-241 by bombarding the Pu-239 nucleus with two neutrons; Pu-241 is an intermediate. Show the complete equation for the synthesis of Am-241.

Solution:

${}_{\mathrm{ 94}}^{239}\text{Pu}$  +  2 ${}_0^1\text{n}$ ---> ${}_{\mathrm{ 94}}^{241}\text{Pu}$ ---> ${}_{\mathrm{ 95}}^{241}\text{Am}$  +  ${}_{\mathrm{-1}}^{\mathrm{ 0}}\text{e}$

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