The overall equation for the fission of235U is235U n right

The overall equation for the fission of^235U is:^235U + n right arrow^236U right arrow^140Ce +^94Zr + 2n The atomic particle masses are:^235U 235.0439 u,^140Ce 139.9054 u,^94Zr 93.9063 u, n 1.00867 u. How much energy, in eV, is released by the disintegration of one atom of^235U How much energy in Joules, is released by the disintegration of one kg of^235U How long will this keep a 100 W light bulb illuminated? (lyr = 3.156 x 10^7s)

Solution

Mass defect dm =(mass of U235)+(mass of neutron) -(mass of Ce 140)-(mass of Zr 94) -2(mass of neutron)

                        = 235.0439u+1.00867 u -139.9054 u- 93.9063u -2(1.00867u)

                        = 0.22353 u

We know 1 u = 931.5 MeV

So, energy released E = dm (931.5 MeV )

                                 = 0.22353 x931.5 MeV

                                 = 208.21 MeV

                                 = 208.21 x10 ⁶ eV

(b).mass m = 1 kg= 1000 g

molecular mass of U 235 is M = 235 g

Number of moles n = m/M

                            = 4.2553 mol

We know 1 mole contain 6.023 x10 ²³ atoms

So, n moles have N = nx6.023 x10 ²³ atoms

                             = 2.5629 x10 ²⁴ atoms

Energy released when N atoms disintegrates is E \' = NE

       E \' = (2.5629 x10 ²⁴ )(208.21 x10 ⁶ eV)

           = 5.3363 x10 ³² eV

           = 5.3363 x10 ³² x1.6 x10 ^-19 J

               =8.538 x10 ¹³ J

(c).Power P = 100 W

Required time t =E \' / P

                      = (8.538 x10 ¹³ ) / 100

                      = 8.538 x10 ¹¹ s

                      =(8.538 x10 ¹¹)/(3.156 x10 ⁷ ) yrs

                      = 27.05388 x10 ³ yrs