# Write a nuclear reaction for the neutron-induced fission of U−235 to produce Te−137 and Zr−97. please help?

May 31, 2015

When uranium-235 undergoes a neutron-induced fission reaction, it will split into two smaller nuclei, which, in your case, are tellurium-137, $\text{^137"Te}$, and zirconium-97, $\text{^97"Zr}$, releasing neutrons and gamma rays in the process.

So, you know that the uranium-235 nucleus must collide with a neutron, which is a subatomic paticle that has no electric charge and a mass approximately equal to that of a proton, 1.

Isotopes are written in the form

$\text{_color(red)(Z)^color(blue)(A)"X}$, where

$\textcolor{red}{Z}$ - atomic number - the number of protons in the nucleus;
$\textcolor{b l u e}{A}$ - mass number - the number of protons + neutrons in the nucleus;
$\text{X}$ - the symbol of the isotope.

To find the atomic number of each isotope that takes part in the reaction, take a look at a periodic table. Your three isotopes will have

• Uranium-235 $\to$ $Z = 92$
• Tellurium-137 $\to$ $Z = 52$
• Zirconium-97 $\to$ $Z = 40$

Now you have what you need to write the nuclear equation.

$\text{_92^235"U" + ""_0^1"n" -> ""_52^137"Te" + ""_40^97"Zr" + color(red)(?)""_0^1"n} + \gamma$

You can figure out how many neutrons the reaction will release by balancing the mass numbers. On the reactants' side you have 235 + 1, and on the products' side you have 137 + 97 + ?.

This means that

"235 + 1" = 137 + 97 + color(red)(?)*1 => color(red)(?) = 236 - 234 = 2

The atomic numbers are already balanced, since you have 92 + 0 on the reactants' side, and 52 + 40 on the products' side.

Therefore, the balanced nuclear equation for this reaction is

$\text{_92^235"U" + ""_0^1"n" -> ""_52^137"Te" + ""_40^97"Zr" + 2""_0^1"n} + \gamma$