zigya tab
A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much U92235 did it contain initially? Assume that the reactor operates 80% of the time that all the energy generated arises from the fission of U92235 and that this nuclide is consumed only by the fission process.

Given, 

Power of reactor  = 1000 MW = 103 MW
                          = 109W
                          = 109 Js-1 

Energy generated by reactor in 5 Years = 5 x 365 x 24 x 60 x 60 x 10

Average energy generated = 200 MeV
                                        = 200 x 1.6 x 10-13 

Number of fission taking place or number of U235 nuclei required, 
                  = 5 × 365 × 24 × 60 × 60 × 109200 × 1.6 × 10-13

                  = 8.2125 × 1026 ×6= 49.275 × 1026 

Mass of 6.023 x 1023 nuclei of U = 235 gm = 235 x 10-3 kg 

Mass of 8.2125 x 1026 nuclei of U,  
             = 235 × 10-36.023 ×1023 × 6 × 8.2125 × 1026 = 1932 kg 
12 of fuel = 1932 kg  Total fuel = 3864 kg.

 

207 Views

Obtain the amount of Co2760 necessary to provide a radioactive source of 8.0 mCi strength. The half-life of Co2760 is 5.3 years.

Given, 
Half life , T = 5.3 years 

Strength of radioactive source,
= 8.0 mCi = 8.0 x 10-3 Ci
= 8.0 x 10-3 x 3.7 x 10-10 disintegrations s-1
= 29.6 x 107 disintegrations s-1
                     
Since the strength of the source decreases with time,
              dNdt = -29.6 × 107

But            dNdt = -λN 

              -λN = -29.6 × 107 

                λN = 29.6 × 107 

                  N = 29.6 × 107λ
                  N = 29.6 × 107× T0.693             λ = 0.693T
                          = 29.6 × 107 × 5.3 × 365 ×24 ×60×600.693 

                        = 7.139 × 1016 

Number of atoms in 60 g of cobalt = 6.023 x  1023 

Therefore,

Mass of 1 atom of cobalt  = 606.023 × 1023g 

Mass of 7.139 × 1016 atoms is, 

= 606.023 × 1023 × 7.139 × 1016g 

= 7.11 μg.                                             

171 Views

The fission properties of Pu94239 are very similar to those of U92235. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure Pu94239 undergo fission?

Given, 
Average amount of energy released per fission,    Pu94239 = 180 MeV 

Quantity of fissionable material  = 1 kg 

In 239 gm Pu, number of fissionable atom or nuclei = 6.023×1023 

In 1 g of Pu, number of fissionable atom or nuclei = 6.023×1023239 

In 1000 gm of Pu, number of fissionable atom or nuclei, 
= 6.023 × 1023239 × 1000= 25.2 × 1023 

Therefore,
Total energy released in fission of 25.2 x 1023 Pu nucleus or in fission of 1 kg pure Pu is, 
                                   = 180 x 25.2 x 1023
                                   
= 4536 x 1023 MeV
                                   = 4.5 x 1026 MeV.

321 Views

Advertisement

How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as:
H12 + H12   He23+n+3.2 MeV.


Power of the electric lamp = 100 W 

When two nuclei of deuterium fuse together, energy released = 3.2 MeV 

Number of deuterium atoms in 2 kg is, 
           = 6.023 × 10232×2000 = 6.023 × 1026

Energy released when 6.023 × 1026 nuclei of deuterium fuse together, 

               = 3.22×6.023×1026 MeV= 3.22×6.023 × 1026 ×1.6 × 10-13J = 1.54 × 1014 J or Ws 
If the lamp glows for time t, then the electrical energy consumed by the lamp is 100 t, 

 100 t =  1.54 × 1014    

            t = 1.54 × 1012s

             = 1.54 × 10123.154 ×107years  = 4.88 × 104 years.

which is the life span of an electric lamp. 

195 Views

Advertisement
Find the Q-value and the kinetic energy of the emitted α-particle in α-decay of (a) Ra88226 and (b) Rn86220
Given   mRa88226 = 226.02540 u;   mRn86222 = 222.01750 u;mRn86220 = 220.01137 u;   mPo84216 = 216.00189 u. 



(a) The reaction invoved is,  

                   Ra88226  Rn86222 + He24

The difference in mass between the original nucleus and the decay products = 226.02540 u - (222.01750 u  + 4.00260 u)
                                    = + 0.0053 u 

 Energy equivalent or Q-value  = 0.0053 x 931.5 MeV 

                                                  = 4.93695 MeV 

                                                  = 4.94 MeV 

The decay products would emerge with total kinetic energy 4.94 MeV.

Momentum is conserved. If the parent nucleus is at rest, the daughter and the α-particle have momenta of equal magnitude p but, in the opposite direction.

Kinetic energy,  K = p22m. 

Since, p is the same for the two particles therefore the kinetic energy divides inversely as their masses.

The α-particle gets 222222+4 of the total i.e., 222226×4.94 MeV or 4.85 MeV. 


(b) The difference in mass between the original nucleus and the decay products = 220.01137 u - (216.00189 u + 4.00260 u)
                                                = 0.00688 u
 Q-value or Energy equivalent = 0.00688 x 931.5 MeV
                                                 = 6.41 MeV
Energy of the alpha particle, Eα= 216216+4×6.41 MeV = 6.29 MeV. 

375 Views

Advertisement