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Let the amount of energy to be produced using nuclear power per year in 2020 is E      (Only 10% of the required electrical energy is to be produced by Nuclear power and only 25% of                                                                                                         thermo-nuclear is successfully converted into electrical energy) Amount of Uranium required to produce this...
(a)   The above fusion reaction releases the energy of 26 MeV Number of Hydrogen atoms in 1.0 kg of Hydrogen is 1000NA Therefore 250NA such reactions would take place The energy released in the whole process is E1 (b) The energy released in fission of one  atom is 200 MeV Number of  atoms present in 1 kg of  is N The energy released on fission of N atoms is E2
decays from 1.088 MeV to 0 V Frequency of  is  Plank's constant, h=6.6210-34 Js  Similarly, we can calculate frequencies of  and  The energy of the highest level would be equal to the energy released after the decay Mass defect is We know 1u = 931.5 MeV/c2 Q value= 0.001473931.5=1.3721 MeV The maximum Kinetic energy of  would be 1.3721-1.088=0.2841 MeV The maximum Kinetic energy of  would be...
To initiate the reaction both the nuclei would have to come in contact with each other. Just before the reaction the distance between their centres would be 4.0 fm. The electrostatic potential energy of the system at that point would be The same amount of Kinetic Energy K would be required to overcome the electrostatic forces of repulsion to initiate the reaction It is given that  Therefore...
The mass defect of the reaction is  1u = 931.5 MeV/c2 Q=0.018883931.5=17.59 MeV
The fission reaction given in the question can be written as The mass defect for the above reaction would be In the above equation, mN represents nuclear masses but 1u =931.5 MeV/c2 Q=0.247995931.5 Q=231.007 MeV Q value of the fission process is 231.007 MeV
1 u = 931.5 MeV/c2 Q=0.03419931.5 =31.848 MeV As the Q value is positive the reaction is energetically allowed 1 u = 931.5 MeV/c2 Q=0.00642931.5 =5.98 MeV As the Q value is positive the reaction is energetically allowed
Let initially there be N1 atoms of  and N2 atoms of  and let their  decay constants be  and  respectively Since initially the activity of   is 1/9 times that of  we have       (i) Let after time t the activity of   be 9 times that of      (ii) Dividing equation (ii) by (i) and taking the natural log of both sides we get where  and t comes out to be 208.5 days
The reaction showing the neutron separation is But 1u=931.5 MeV/c2 Therefore E=(0.014019)931.5 E=13.059 MeV Therefore to remove a neutron from the  nucleus 13.059 MeV of energy is required
The reaction showing the neutron separation is But 1u=931.5 MeV/c2 Therefore E=(0.008978)931.5 E=8.363007 MeV Therefore to remove a neutron from the  nucleus 8.363007 MeV of energy is required
Let the abundances of  and  be x and y respectively. x+y+78.99=100 y=21.01-x The average atomic mass of Mg is 24.312 u The abundances of  and   are 9.3% and 11.71% respectively
For the electron capture, the reaction would be The mass defect and q value of the above reaction would be where mN and mN are the nuclear masses of elements X and Y respectively For positron emission, the reaction would be The mass defect and q value for the above reaction would be From the above values, we can see that if Q2 is positive Q1 will also be positive but Q1 being positive does...
Mass of an element with mass number A will be about A u. The density of its nucleus, therefore, would be As we can see the above density comes out to be independent of mass number A and R0 is constant, so  matter density is nearly constant
For a head-on collision of two deuterons, the closest  distances between their centres will be d=2r d=22.0 d=4.0 fm d=410-15 m charge on each deuteron = charge of one proton=q =1.610-19 C The maximum electrostatic potential energy of the system during the head-on collision will be E The above basically means to bring two deuterons from infinity to each other would require 360 keV of work to be...
The energy liberated on the fusion of two atoms of deuterium= 3.27 MeV Number of fusion reactions in 2 kg of deuterium = NA500 The energy liberated by fusion of 2.0 kg of deuterium atoms E   Power of lamp (P)= 100 W Time the lamp would glow using E amount of energy is T= =4.99104 years
The amount of energy liberated on fission of 1   atom is 200 MeV. The amount of energy liberated on fission of 1g    Total Energy produced in the reactor in 5 years Mass  of  which underwent fission, m =1537.8 kg The amount present initially in the reactor = 2m =21537.8 =3075.6 kg
Number of atoms present in 1 kg(w) of   =n Energy per fission (E)=180 MeV Total Energy released if all the atoms in 1 kg   undergo fission = E  n =180  2.521024 =4.5361026 MeV
The reaction will be  The mass defect of the reaction will be Since the mass defect is negative the Q value will also negative and therefore the fission is not energetically possible.
The above positive value of mass defect implies Q value would be positive and therefore the reaction is exothermic.
The above negative value of mass defect implies there will be a negative Q value and therefore the reaction is endothermic
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