Calculate mass loss in fission of Uranium-235

[xatu]

Problem:

The fission of [itex]^{235}_{92}U[/itex] releases approximately 200 MeV. What percentages of the orginal mass of [itex]^{235}_{92}U[/itex] + n disappears?

Solution:

Using E=mc^2 we can estimate the amount of mass m converted into energy E. Solving for m we get,

m=E/c^2=[(2 x 10^8 eV)(1.60 x 10^-19 J/eV)]/[3 x 10^8 m/s]^2=3.56 x 10-28 kg

Certainly a small amount. However, the answer in my text is 0.1%. How can this mass be converted into a percentage when the initial quantity was never given?

 

[TSny]

However, the answer in my text is 0.1%. How can this mass be converted into a percentage when the initial quantity was never given?

You'll need to calculate the approximate mass of the U-235 nucleus.

 

[xatu]

That makes sense. I got the right answer, too.

Total mass is ≈ (1.67 x 10^-27 kg)*235 = 3.92 x 10^-25 kg. So the fraction of mass converted into energy is (3.56 x 10^-28 kg)/(3.92 x 10^-25)*100% ≈ 0.1%.

Thanks.