A Brief Tutorial About Writing Nuclear Symbols

**Extremely Brief Historical Overview**

The neutron was discovered in 1932 by James Chadwick. More or less immediately, the neutron was seen as a tool to probe the nucleus of the atom and work commenced in many places.

One of those places was in Rome, the work being headed up by Enrico Fermi. About 1934, he thought he had discovered new elements beyond uranium, however he had fission take place, but did not recognize it as such.

The discovery of nuclear fission took place in early December 1938, in Berlin, by Otto Hahn and Fritz Strassmann. The explanation for the physical mechanism of fission took place over the Christmas holiday of 1938, in Stockholm. This work was carried out by Lise Meitner and Otto Frisch. Meitner, who had work closely with Hahn on fission, had been forced to flee Nazi Germany in July 1938 and had found a job in Stockholm.

Obviously, work relevant to the discovery of fission took place before 1932 and fission continued to be studied after 1938. My goal here was only to give you a brief look at the relevant events. You can read a bit more about the discovery of fission here and here.

A Brief Tutorial About Writing Nuclear Symbols

**The Fission Reaction**

The first three fission equations all use isotopes of barium and isotopes of krypton for the products. I did that to demonstrate that more than one isotope of a given element can show up as a fission product. I do not want you to draw the conclusion that only the elements barium and krypton show up as fission products.

The first ten examples are different fission reactions with some comments attached here and there. Examples that must be solved start at #11.

**Example #1:** Here is a typical fission equation:

${\text{}}_{92}^{235}\text{U}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{56}^{140}\text{Ba}$ + ${\text{}}_{36}^{94}\text{Kr}$ + 2 ${\text{}}_{0}^{1}\text{n}$ + QQ stands for the nuclear energy produced. Sometimes, γ (Greek letter gamma) is used. Sometimes the word energy is used and you will also see MeV (megaelectron volts) used. MeV is a common measure of energy used in nuclear reactions.

**Example #2:** Often, in the writing of a fission equation, the energy (usually in the form of gamma radiation) released is ignored. Like this:

${\text{}}_{92}^{235}\text{U}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{56}^{143}\text{Ba}$ + ${\text{}}_{36}^{90}\text{Kr}$ + 3 ${\text{}}_{0}^{1}\text{n}$I plan to ignore the energy term in much of what I write below. Energy will be involved in some examples below, starting with #14.

**Example #3:** Sometimes, the nuclide that is formed for a brief instant before fissioning, is shown:

${\text{}}_{92}^{235}\text{U}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{92}^{236}\text{U}$ ---> ${\text{}}_{56}^{141}\text{Ba}$ + ${\text{}}_{36}^{92}\text{Kr}$ + 3 ${\text{}}_{0}^{1}\text{n}$

**Two Important Rules About Writing Fission Reactions**

I. The atomic numbers have to add up to be equal on each side of the arrow.

In all three cases above, the total atomic number on the left-hand side is 92. On the right-hand side of the first equation, we have this:56 + 36 + 0 + 0 = 92Here's the second and third:

56 + 36 + 0 + 0 + 0 = 9256 + 36 + 0 + 0 + 0 = 92

I put zeros in there to remind you about the neutrons.

II. The mass numbers have to add up to be equal on each side of the arrow.

In all three cases above, the total mass number on the left-hand side is 236. On the right-hand side of the first equation, we have this:140 + 94 + 1 + 1 = 236Here's the second and third:

143 + 90 + 1 + 1 + 1 = 236141 + 92 + 1 + 1 + 1 = 236

I put ones in there to remind you about the neutrons.

**Example #4:** Here is another fission reaction:

${\text{}}_{92}^{233}\text{U}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{51}^{133}\text{Sb}$ + ${\text{}}_{41}^{98}\text{Nb}$ + 3 ${\text{}}_{0}^{1}\text{n}$

**Example #5:** And another:

${\text{}}_{94}^{239}\text{Pu}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{58}^{144}\text{Ce}$ + ${\text{}}_{36}^{94}\text{Kr}$ + 2 ${\text{}}_{0}^{1}\text{n}$The most common nuclide used for fission reaction examples is ${\text{}}_{92}^{235}\text{U}$, but you do see the two just above from time to time. There's also a ${\text{}}_{92}^{236}\text{U}$ example below (at #10).

**Example #6:** Here's another ${\text{}}_{92}^{233}\text{U}$ example:

${\text{}}_{92}^{233}\text{U}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{92}^{234}\text{U}$ ---> ${\text{}}_{54}^{137}\text{Xe}$ + ${\text{}}_{38}^{94}\text{Sr}$ + 3 ${\text{}}_{0}^{1}\text{n}$

**Example #7:** And one more ${\text{}}_{94}^{239}\text{Pu}$
example:

${\text{}}_{94}^{239}\text{Pu}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{94}^{240}\text{Pu}$ ---> ${\text{}}_{54}^{134}\text{Xe}$ + ${\text{}}_{40}^{103}\text{Zr}$ + 3 ${\text{}}_{0}^{1}\text{n}$

**Example #8:** Here's a wild one:

${\text{}}_{92}^{235}\text{U}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{61}^{151}\text{Pm}$ + ${\text{}}_{35}^{82}\text{Br}$ + 3 ${\text{}}_{0}^{1}\text{n}$ + 4β¯That β¯ stands for a beta particle, which is known to be an electron. When you total up the atomic numbers, use −1 for the electron's "atomic number" and 0 for the mass number. The atomic number calculation looks like this:

92 + 0 = 35 + 61 + 0 + (−4)and the mass number calculation looks like this:

235 + 1 = 151 + 82 + 3 + 0

For these last two examples, I will use alternate formatting styles.

**Example #9:** My preferred alternate formatting style:

92-U-235 + 0-n-1 ---> 57-La-148 + 35-Br-85 + 3(0-n-1)

**Example #10:** A formatting style I have seen used:

236/92 U + 1/0 n ---> 100/42 Mo + 126/50 Sn + 11(1/0 n)Personally, I don't care too much for this formatting style, but you do see it used on the Internet a fair amount, especially in questions and answers on Yahoo Answers. That means you have to know it.

This example uses the fission of U-236, a nuclide not often used in fission examples. Another nuclide that can undergo fission, but is not often seen is Pu-241.

Sometimes the slash is replaced with a dash:

235-92 U + 1-0 n ---> 95-42 Mo + 139-50 Sn + 2(1-0) n

**Example #11:** An atom of uranium-235 undergoes fission when bombarded with a neutron to yield an atom of zirconium-97, two neutrons, and another unknown product. Identify the nuclear symbol for the unknown product.

**Solution:**

1) Write as much of the fission reaction as possible:

235/92 U + 1/0 n ---> 97/40 Zr + 2(1/0 n) +_____

2) The atomic numbers must total up to be the same on each side of the arrow. This holds true for the mass numbers as well.

The atomic number of the product must be:92 − 40 = 52The mass number of the product must be:

235 + 1 − 97 − 2 = 137

3) The element with an atomic number of 52 is Te. So, the other product must be:

137/52 TeNote: sometimes, the space is not used, as in 137/52Te

**Example #12:** A nucleus of uranium-235 undergoes fission when it absorbs a neutron. An atom of barium-139, three neutrons, and another product are the results. Determine the unknown product and write the full fission reaction.

**Solution:**

1) Determine the atomic number of the unknown product:

The atomic number for barium is 56. The total atomic number on the left-hand side is 92.92 − 56 = 36

That's the atomic number of krypton.

2) Determine the mass number of the unknown product:

The mass number for Ba is 139 and three neutrons are released. That makes for 142. The total mass number on the left-hand side is 236.236 − 142 = 94. That's the mass number of the other product.

36-Kr-94

3) The full reaction:

92-U-235 + 0-n-1 ---> 92-U-236 ---> 56-Ba-139 + 36-Kr-94 + 3(0-n-1)

**Example #13:** What is the missing term X in the equation?

235/92 U + 1/0 n ---> 99/41 Nb + X + 4 (1/0 n)

**Solution:**

1) Total atomic and mass numbers of left-hand side:

92 + 0 = 92

235 + 1 = 236

2) Determine missing atomic number:

92 − 41 = 51

3) Determine missing mass number:

236 − 99 − 4 = 133

4) The answer:

133/51 Sb

**Example #14:** Write the isotope needed, by identifying A, Z, and X, to balance the following nuclear fission reaction:

${\text{}}_{92}^{235}\text{U}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{54}^{140}\text{Xe}$ + ${\text{}}_{\mathrm{Z}}^{\mathrm{A}}\text{X}$ + 4 ${\text{}}_{0}^{1}\text{n}$

**Solution:**

atomic number ---> 92 + 0 = 54 + Z + 0 ---> Z = 38

mass number ---> 235 + 1 = 140 + A + 4 ---> A = 92results in:

${\text{}}_{92}^{235}\text{U}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{54}^{140}\text{Xe}$ + ${\text{}}_{38}^{92}\text{Sr}$ + 4 ${\text{}}_{0}^{1}\text{n}$

**Example #15:** Calculate the energy released when the following fission reaction occurs:

${\text{}}_{92}^{235}\text{U}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{56}^{141}\text{Ba}$ + ${\text{}}_{36}^{92}\text{Kr}$ + 3 ${\text{}}_{0}^{1}\text{n}$ + γ

**Discussion:**

Some of the mass at the start (the U-235 and the neutron) will disappear during the fission. We have to subtract the total mass on the right-hand side from the total mass on the left-hand side. We will then use Einstein's mass-energy relation (E = mc^{2}) to calculate the energy generated in the fission. Report the value in MeV (megaelectron volts).By the way, electrons are completely ignored. This is because the number of electrons remains constant on both sides of the reaction arrow.

**Solution:**

1) Since the text of the example provides no data whatsoever, we must do some looking up. I have decided to start with the masses in unified atomic mass units:

Nuclide Mass (u) Nuclide Mass (u) ${\text{}}_{92}^{235}\text{U}$ 235.04393 ${\text{}}_{56}^{141}\text{Ba}$ 140.914411 ${\text{}}_{0}^{1}\text{n}$ 1.008665 ${\text{}}_{36}^{92}\text{Kr}$ 91.926156

2) Left-hand side:

235.04393 + 1.008665 = 236.052595 u

3) Right-hand side:

140.914411 + 91.926156 + (3)(1.008665) = 235.866562 u

4) The difference (known as the mass defect):

236.052595 − 235.866562 = 0.186033 uFrom here, we can go a long way (steps 5 to 8) or a short way (step 9). It all depends on what you are given and what the teacher wants.

5) We will next convert the mass defect to kilograms, so back to looking things up:

1 u = 1.6605402 x 10¯^{27}kg (source)

6) Convert u to kg:

(0.186033 u) (1.6605402 x 10¯^{27}kg / u) = 3.08915275 x 10¯^{28}kg

7) Use Einstein's mass-energy conversion:

E = mc^{2}E = (3.08915275 x 10¯

^{28}kg) (2.99792458 x 10^{8}m/s)^{2}E = 2.7764 x 10¯

^{11}JI used Joules because the unit kg-m

^{2}/s^{2}is the unit for Joules.

8) Last calculation is to convert J to MeV.

1 J = 6241506479963.2 MeV (source)(2.7764 x 10¯

^{11}J) (6241506479963.2 MeV / J) = 173.3 MeV

9) This is the shorter way:

1 u = 931.494893 MeV / c^{2}(calculated here)(0.186033 u) (931.494893 MeV / u-c

^{2}) = 173.3 MeV/c^{2}

**Example #16:** Calculate the energy released (in MeV) when the following fission reaction occurs:

${\text{}}_{92}^{235}\text{U}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{55}^{137}\text{Cs}$ + ${\text{}}_{37}^{95}\text{Rb}$ + 4 ${\text{}}_{0}^{1}\text{n}$ + γ

**Discussion:**

The plan of attack is based on this:

1 u = 931.494893 MeV/c^{2}= 0.931494893 GeV/c^{2}

That's the mass, the energy is this:

E = mc^{2}E = (0.931494893 GeV/c

^{2}) (c^{2})E = 0.931494893 GeV

**Solution:**

1) Determine the energy (in GeV) for each nuclide involved.

U-235 ---> (235.04393 u) (0.931494893 GeV/u) = 218.94222 GeVCs-137 ---> (136.90709 u) (0.931494893 GeV/u) = 127.528255 GeV

Rb-95 ---> (94.929303 u) (0.931494893 GeV/u) = 88.426161 GeV

three neutrons ---> (3) (1.008665 u) (0.931494893 GeV/u) = 2.8186989 GeV

I eliminated one neutron common to both sides. I did not do that in example #14.

2) Subtract:

218.94222 − (2.8186989 + 127.528255 + 88.426161) = 0.1691051 GeV0.1691051 GeV = 169.1 MeV

**Example #17:** Calculate the energy produced for this fission:

${\text{}}_{92}^{238}\text{U}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{38}^{96}\text{Sr}$ + ${\text{}}_{54}^{140}\text{Xe}$ + 3 ${\text{}}_{0}^{1}\text{n}$ + γ

**Solution:**

E = mc^{2}E = (m

_{U-238}− m_{Sr-96}− m_{Xe-140}− 2n) (c^{2})E = (238.050784 u − 95.921750 u − 139.92164 u − (2) (1.008665 u)) (c

^{2})E = 0.190064 u-c

^{2}Remembering that 1 u = 931.5 MeV/c

^{2}, we find:E = (0.190064 u-c

^{2}) (931.5 MeV/u-c^{2})E = 177.0 MeV

**Example #18:** Isotopes of barium and krypton are quite commonly seen in fission examples. Several were written above. Here are two more:

92-U-235 + 0-n-1 ---> 56-Ba-142 + 36-Kr-91 + 3(0-n-1)${\text{}}_{92}^{235}\text{U}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{56}^{138}\text{Ba}$ + ${\text{}}_{36}^{95}\text{Kr}$ + 3 ${\text{}}_{0}^{1}\text{n}$

**Example #19:** The most common nuclide in fission examples is U-235. Above were some examples of other nuclides that fission. Here is one involving U-234:

${\text{}}_{92}^{234}\text{U}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{51}^{133}\text{Sb}$ + ${\text{}}_{41}^{98}\text{Nb}$ + 3 ${\text{}}_{0}^{1}\text{n}$

**Example #20:** Example #8 was a fission that included some beta particles being emitted. Here's another:

${\text{}}_{92}^{235}\text{U}$ + ${\text{}}_{0}^{1}\text{n}$ ---> ${\text{}}_{58}^{140}\text{Ce}$ + ${\text{}}_{40}^{94}\text{Zr}$ + 2 ${\text{}}_{0}^{1}\text{n}$ + 6β¯