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

1 Answer
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, #""^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#