Answer:
Energy released = [tex]2.25\times 10^{14}\;J[/tex]
Explanation:
Amount of fissionable material = 2.5 kg
Mass defect = 0.10 % of fissionable material
Mass defect [tex]=2.5\times \frac{0.1}{100} = 0.0025\; kg[/tex]
[tex]E = mc^{2}[/tex]
Where,
m = Mass defect in Kg
C = Speed of light m/s2
[tex]C = 3\times 10^{8}\;m/s^2[/tex]
Now substitute the values in the above equation
[tex]E = 0.0025\times (3\times 10^{8})^2 = 2.25\times 10^{14}\;J[/tex]