Question

How many grams of $U^{235}$ is required to produce $8\times {10}^{11}J$ of energy, if $200MeV$ of energy is released when one $U^{235}$ atom undergoes fission?

[Assume every atom disintegrates in the process]

• A. $39.54g$
• B. $9.754g$
• C. $19.51g$
• D. $49.754g$

Answer 1

The correct option is B $9.754g$

$8\times {10}^{11}J=8\times {10}^{11}\times 6.242\times {10}^{12}MeV=49.9\times {10}^{23}MeV$

200MeV is released for one atom.

So, no. of atom $=\frac{49.9\times {10}^{23}}{200}=0.24\times {10}^{23}atom$

So, No. of moles $=\frac{0.24\times {10}^{23}}{NA}=0.041molSo,mass=0.041\times 235=9.754g$