Question

Find the energy released in the following reaction:
n + ²³⁵U ? ¹⁴¹Ba + ⁹²Kr + 3n Use the following atomic mass table to determine your answers.

Answer 1

energy released = ?mc2

?m = {( 235.0439231 + 1.008665 ) - ( 140.922940+ 91.919524+ 3 * 1.008665) }

      = 0.1841291

energy released = ?m c2 = 0.1841291 amu * C2

                        = 0.1841291 * 931.5 Mev                    since 1 amu * c2 = 931.5 Mev

                        =171.516 Mev

Answer 2

deltaM = 235.0439 + 1.008665 - 140.9144 - 91.9262 -3* (1.008665)

           = 0.1860 u

energy released

E = deltaM*931.5

   = 0.1860*931.5

= 173 MeV

= 2.77*10^11 J

Answer 3