Okay, let's solve this step-by-step:
Given:
- Uranium-235 undergoes fission
- Energy released per reaction = 200 MeV
- Mass of U-235 = 235 amu
To find: Balanced nuclear fission reaction
Analysis:
- In nuclear fission, a heavy nucleus like U-235 splits into two lighter nuclei and neutrons are released.
- The missing mass gets converted to energy as per Einstein's mass-energy equivalence equation E=mc^2.
Calculating mass deficit:
- 1 amu = 1.66 x 10-27 kg
- So 235 amu = 235 x 1.66 x 10-27 kg = 3.9 x 10-25 kg
- With 200 MeV energy released:
E = mc^2
200 x 106 eV = (m)c2
m = E/c2 = (200 x 106 x 1.6 x 10-19)/((3 x 108)2)
= 2.2 x 10-30 kg
This is the mass deficit in the reaction.
Balanced Reaction:
Let's assume the products are Kr and Ba.
Total mass number must be conserved:
235 = A(Kr) + A(Ba)
Let A(Kr) = 90 and A(Ba) = 140
The atomic numbers must also match:
92 = Z(Kr) + Z(Ba)
Z(Kr) = 36 and Z(Ba) = 56
The complete balanced reaction is:
235U → 90Kr + 140Ba + 3n
So the nuclear fission reaction for Uranium-235 is:
235U + n → 90Kr + 140Ba + 3n