Engineering Chemistry S Datta

NUCLEAR CHEMISTRY 49
So, the mass lost is 0.2153 a.m.u., which is equivalent to an energy of 0.2153 × 931
≈ 200 MeV.
Highlight:
It can be shown that when 1 kg of uranium is completely fissioned, the energy released
would be 8.2 × 10 13 J, which is sufficient to supply energy at the rate of 2.2 megawatt
continuously for one year.
To utilise the energy released during the fission, we are to control the chain reaction to
get the heat energy according to necessity. Now-a-days, in an atomic reactor, the motion of
neutrons are retarded by graphite or heavy water to control the chain reaction. So, fission
reaction is retarded and thereby controlled and the heat generated is being absorbed by molten
Na–K alloy and this heat is utilised in generating steam for thermal power.
Atomic Fusion
At a very high temperature, two nuclei combine to give a comparatively heavy nucleus,
i.e., two nuclei combine to give a new atom. This phenomenon is known as atomic fusion. And
during this fusion reaction some mass is destroyed, that means such a fusion process would
also lead to liberation of huge amount of energy which causes explosion.
∴
2
1
2
1
4
2
H + H ⎯→ He
+ Energy
2 × 2.0162 ⎯→ 4.00389
∆m = 0.02863 a.m.u.
which is equivalent to 0.02863 × 931 MeV
= 26.6545 MeV. (Energy released due to above fusion)
The difficulty is to attain the high temperature for such fusion process. But it is believed
that such condition is attained for preparing thermonuclear bombs such as hydrogen bomb.
The transformations for fusion reactions are:
(i) 1 3 H+
(ii) 1 3 H+
2
1 H ⎯→ 4 2 He + 1 0 n + 17 MeV
1
1 H⎯→ 4 2 He + 20 MeV etc.
Mass Defect and Binding Energy
After the discovery of isotopes, Aston examined the atomic weights of various elements
and concluded that the atomic weights of all the elements would be a whole number. But due
to the presence of isotopes in unequal amounts, the atomic weights become fractions. Besides
that, the atomic weights of the different elements depend upon another factor: in a.m.u. the
atomic weights are calculated from number of protons, electrons and neutrons. The calculated
mass (M) and actual weight (A) is determined considering the contribution for the presence of
isotopes. The difference of these two quantities is called mass defect (∆) or ∆m
∴ ∆ = M – A.
For a 3 7 Li atom, the atomic weight will be
3 × 0.0005486 + 3 × 1.00758 + 4 × 1.00893
= 7.0601058 a.m.u.
But actual atomic weight of 3 7 Li is 7.01818 a.m.u. (in 16 O scale)
So, mass defect (∆) = 7.0601058 – 7.01818
= 0.0419258 a.m.u.