The rating of a bomb is magnitude of energy released, specified in terms of the mass of TNT required to produce the same energy. One megaton (10 6 ton) of TNT releases 2.6 × 10 28 MeV of energy. An early proposal for a hydrogen bomb consists of deuterium ( 
) as fuel. The high temperature and particle density needed for fusion are provided by an atom bomb (trigger) which involves a U 235 fission fuel that is arranged to impress an imploding compressive shock wave on deuterium. The operative fusion reaction is
5 2 H → 3 He + 4 He + 1 H + 2n
The atomic masses and particle masses needed to answer the following questions are:
1 H 1.007825 u
4 He 4.002603 u
2 H 2.014102 u
3 He 3.016029 u
n 1.008665 u
e 0.0005486 u
(i) In the given reaction, the energy is –
Text Solution
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Ans.
(i)
Sol. The given reaction is
5 2 H → 3 He + 4 He+ 1 H +2n
The mass defect is mass of reactant nuclei minus the mass of product nuclei
Δ m = 
– { 
+ 
+ 
+ 2m n }
Here all masses are nuclear masses (not atomic masses).
Let us represent the atomic masses by capital M's
⇒ Δ m = 5{ 
–2m e } - { 
+ 
+ M H + 2m n – 8m e }
5 
– { 
+ 
+ M H + 2m n } – 2m e
= 5 × 2.01 × 102 – {3.01602 + 4.002603 + 1.007825 + 2.019330} – 2 × m e
= 0.026723 – 2m e
Q = (0.026723 × 931 – 0.5 × 2) MeV
(ii)
Sol. The given reaction is
5 2 H → 3 He + 4 He+ 1 H +2n
The mass defect is mass of reactant nuclei minus the mass of product nuclei
Δ m = 
– { 
+ 
+ 
+ 2m n }
Here all masses are nuclear masses (not atomic masses).
Let us represent the atomic masses by capital M's
⇒ Δ m = 5{ 
–2m e }- { 
+ 
+ M H + 2m n – 8m e }
5 
– { 
+ 
+ M H + 2m n } – 2m e
= 5 × 2.01 × 102 – {3.01602 + 4.002603 + 1.007825 + 2.019330} – 2 × m e
= 0.026723 – 2m e
Q = (0.026723 × 931 – 0.5 × 2) MeV
Since m e = 0.5 MeV/c 2
Q = 23.8 MeV
(iii)
Sol. Now 500 kg of deuterium will have
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