Nuclear Fission

Energy in Nuclear Fissions

Nuclear reactions (fission)Otto Hahn and Frederic Strassmann managed to bombard a nucleus of with neutrons (1938). The product isotopes of this reaction were radioactive and they thought they were turning the uranium nuclei into heavier nuclei. Instead, when they analysed the product isotopes chemically, they discovered that the nuclei had split into lighter radioactive isotopes of barium and krypton.

U23592

Incoming neutron

U-235 nucleus

Kr-90 nucleus

Ba-144 nucleus

Nuclear fissionThe word equation of this particular nuclear fission is represented underneath:

releasedQenergynKrBanU __2 109036

14456

10

23592

What is the total number of nucleons before and after the reaction?

a) Less than before

b) More than before

c) The same as before

Nuclear fission

releasedQenergynKrBanU __2 109036

14456

10

23592

The word equation of this particular nuclear fission is represented underneath:

What is the total number of nucleons before and after the reaction?

a) Less than before

b) More than before

c) The same as before

Nuclear fissionOther experiments showed that:

• Several neutrons are released in the fission

• Different products of the fission reaction can occur, but the number of protons and neutrons and the total energy is conserved.

• The products are radioactive. In fact, they normally have a greater neutron/proton ratio than the stable nuclei of the same elements

• Slow neutrons achieve nuclear fission of U-235 more easily than fast ones. This is because of the potential well around a nucleon.

• The energy released in nuclear fission is much higher than energy released by chemical processes.

Nuclear fission

releasedQenergynKrBanU __2 109036

14456

10

23592

Calculate the energy released Q in the following nuclear reaction.

For your calculations click on the link below to find out the nuclear masses.

Nuclear mass finder

m(U-235) =

m(n) =

m(Ba-144) =

m(Kr-90) =

2m(n) =

143.92279 u

235.04438 u

1.00866 u

89.92089 u

2.01732 u

m = tot m (before) – tot mass (after) = 0.1921 u

E equivalence of mass loss = 0.1921 x 931 MeV =

= 178.85 MeV

Energy in the nuclear fissionSo, one nucleus of U-235 gives out about 200 MeV of energy, because the energy released is the same as the energy equivalence of the mass loss in the reaction.

If we consider 6 x 1023 nuclei of U-235, i.e. the number of nuclei in 235 g of Uranium 235, the energy release is 12 x 1023 MeV = 2 x 1013 J.

Calculate the energy released from the fission of 1 kg of U-235

Work out the amount of U-235 needed to run a 100 MW power station for one year, if its efficiency is 35%.

JEE 13

1313

105.81000235

102

1000235

102

Energy in the nuclear fissiont = 60 x 60 x 24 x 365 (s) = 3.1536 x 107 s

Energy produced per year (100% eff) = Power x t = = 108 W x 3.1536 x 107 s ~ 3 x 1015 J

m(U-235) required = 9 x 1015 : 8 x 1013 = 112.5 kg

JeffEE 15109)%100(3%)35(3%35

%100