Question

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

Answer 1

When uranium-235 undergoes a neutron-induced fission reaction, it will split into two smaller nuclei, which, in your case, are tellurium-137, `""^137"Te"`, and zirconium-97, `""^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.

Those will be your rectants.

Isotopes are written in the form

`""_color(red)(Z)^color(blue)(A)"X"`, where

`color(red)(Z)` - atomic number - the number of protons in the nucleus;
`color(blue)(A)` - mass number - the number of protons + neutrons in the nucleus;
`"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 `->` `Z = 92`
• Tellurium-137 `->` `Z=52`
• Zirconium-97 `->` `Z=40`

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

`""_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

`""_92^235"U" + ""_0^1"n" -> ""_52^137"Te" + ""_40^97"Zr" + 2""_0^1"n" + gamma`