Class 12 Physics Nuclei | Nuclear Reactor |

**Nuclear Reactor**

- Nuclear reactors are useful in producing electricity.
- A nuclear reactor is an arrangement to generate electricity whichmakes use of nuclear fission.

__Requirements for controlled nuclear fission in reactor__:

- Neutrons to be slowed down.

- Neutrons are slowed down by using
__moderators__which are lighter nuclei which slow down fast moving neutrons by elastic collision. - Commonly used moderators:-

1.Water

- Heavy Water
- Graphite

- Consequence of use of moderators:-
- Multiplication factor of neutrons increases: - When a neutron hits a target nucleus along with daughter nucleus,it produces 3 neutrons.These 3 neutrons are highly energetic but they need to be slowed down, so they can hit the target nucleus.
- As a result high multiplication factor results in uncontrolled chain reaction.

- Excess neutrons to be absorbed.

- As uncontrolled chain reaction is wanted therefore to absorb excess neutrons
__Control Rods__are used. - These control rods are inserted in the core of the nuclear reactor.
- Control rods are capable of initiating(while taking out of the reactor) and stopping(inserting in the nuclear reactor) the nuclear reaction.
- As they absorb all the excess neutronsthere are no neutrons left to start the reaction.
- Control rods are made up of neutron absorbing materials.
- They decrease the multiplication factor of neutrons to a very small value.
- Commonly used material is
__Cadmium__.

__Control Rods__

__Construction of Nuclear Reactor__

- The core of the nuclear reactor consists of uranium (
^{235}U) in the form of cylindrical rods.These rods are dipped inside a liquid which is the moderator. - Whenever one neutron strikes this uranium rod nuclear fission reaction starts and 3 fast moving neutrons are produced.
- Because of the moderator these 3 neutrons undergo elastic collision as a result they slow down before they strike the second rod.
- Geometry of the core is such that only one out of 3 neutrons which are emitted strike the next rod making the reaction a controlled one.
- When the control rods are inserted inside they will absorb all the extra neutrons.Since there are no neutrons nuclear fission reaction will stop.
- Large amount of energy is also released in the core.
- In order to extract the energy from the core water at very high pressure is passed through it.
- As hot water passes through it produces steam in the steam generators.
- This steam is used to run the turbines which in turn produce electricity.
- This process will keep on continuing till the uranium on the rods does not get over.Then the rods have to be replaced in the nuclear reactor.

** Advantages**:-

- Energy released is extremely large.
- Needs fuel in extremely small quantity.

** Disadvantages**:-

- Spent fuel is highly radioactive and extremely hazardous to all life forms.
- Accumulation of radioactive waste.

** Problem:- **A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much (

** Answer:- **Half-life of the fuel of the fission reactor, t

= 5 × 365 × 24 × 60 × 60 s

We know that in the fission of 1 g of (^{235}_{92}U) nucleus, the energy released is equal to 200MeV.

1 mole, i.e., 235 g of (^{235}_{92}U) contains (6.023 × 10^{23}) atoms.

1 g (^{235}_{92}U) contains = (6.023x10^{23})/ (235) atoms.

The total energy generated per gram of (^{235}_{92}U) is calculated as:

E = ((6.023x10^{23})/ (235)) x200MeV/g

= (200x6.023x10^{23}x1.6x10^{-19}x10^{6})/ (235) =8.20x10^{10} J/g

The reactor operates only 80% of the time.

Hence, the amount of (^{235}_{92}U) consumed in 5 years by the 1000 MW fission reactor iscalculated as:

= (5x80x60x60x365x24x1000x10^{6}) g / (100x8.20x10^{10})

=1538kg

Therefore, Initial amount of (^{235}_{92}U) = 2 × 1538 = 3076 kg

** Problem:- **The fission properties of (

** Answer:- **Average energy released per fission of (

Amount of pure (^{239}_{94}Pu), m = 1 kg = 1000 g

N_{A}= Avogadro number = 6.023 × 10^{23}

Mass number of (^{239}_{94}Pu) = 239 g

1 mole of (^{239}_{94}Pu) contains N_{A} atoms.

Therefore mg of (^{239}_{94}Pu) contains ((N_{A})/ (Mass number) x m) atoms

= ((6.023 × 10^{23})/ (239) x1000) =2.52 x10^{24} atoms.

Therefore Total energy released during the fission of 1 kg of (^{239}_{94}Pu) is calculated as:

E=E_{av} x 2.52x10^{24}

=180x2.52x10^{24} = 4.536x10^{26}MeV

Hence, 4.536x10^{26}MeV is released if all the atoms in 1 kg of pure (^{239}_{94}Pu) undergo fission.

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