The explosive in a hydrogen bomb is a mixture of \( ^2\text{H} \), \( ^3\text{H} \), and \( ^6\text{Li} \) in some condensed form. The chain reaction is given by\[_3^6\text{Li} + _0^1\text{n} \rightarrow _2^4\text{He} + _1^3\text{H}\]\[_1^2\text{H} + _1^3\text{H} \rightarrow _2^4\text{He} + _0^1\text{n}\]During the explosion, the energy released is approximately ____.[Given: \( M(\text{Li}) = 6.01690 \, \text{amu} \), \( M(^3\text{H}) = 2.01471 \, \text{amu} \), \( M(_2^4\text{He}) = 4.00388 \, \text{amu} \), and \( 1 \, \text{amu} = 931.5 \, \text{MeV} \)]
- 28.12 MeV
- 12.64 MeV
- 16.48 MeV
- 22.22 MeV
The Correct Option is D
Approach Solution - 1
To solve this problem, we need to calculate the energy released during the reactions given. The reactions occur as follows:
- The reaction of lithium with a neutron: \(_3^6\text{Li} + _0^1\text{n} \rightarrow _2^4\text{He} + _1^3\text{H}\)
- The reaction of deuterium with tritium: \(_1^2\text{H} + _1^3\text{H} \rightarrow _2^4\text{He} + _0^1\text{n}\)
We are given the following atomic masses:
- \(M(_3^6\text{Li}) = 6.01690 \, \text{amu}\)
- \(M(_1^3\text{H}) = 3.01605 \, \text{amu}\)
- \(M(_2^4\text{He}) = 4.00388 \, \text{amu}\)
- Neutron mass \(= 1.008665 \, \text{amu}\)
- Deuterium mass \(= 2.01471 \, \text{amu}\)
The energy released in a nuclear reaction is calculated using the mass defect, which is the difference between the mass of products and the mass of reactants. This difference is then converted to energy using the relation:
\(\Delta E = \Delta m \cdot 931.5 \, \text{MeV/amu}\)
Calculate for First Reaction:
Reactants: \(_3^6\text{Li} + _0^1\text{n} = 6.01690 + 1.008665 = 7.025565 \, \text{amu}\)
Products: \(_2^4\text{He} + _1^3\text{H} = 4.00388 + 3.01605 = 7.01993 \, \text{amu}\)
Mass defect: \(\Delta m = 7.025565 - 7.01993 = 0.005635 \, \text{amu}\)
Energy released: \(\Delta E_1 = 0.005635 \times 931.5 = 5.25 \, \text{MeV} \, \text{(approximately)}\)
Calculate for Second Reaction:
Reactants: \(_1^2\text{H} + _1^3\text{H} = 2.01471 + 3.01605 = 5.03076 \, \text{amu}\)
Products: \(_2^4\text{He} + _0^1\text{n} = 4.00388 + 1.008665 = 5.012545 \, \text{amu}\)
Mass defect: \(\Delta m = 5.03076 - 5.012545 = 0.018215 \, \text{amu}\)
Energy released: \(\Delta E_2 = 0.018215 \times 931.5 = 16.97 \, \text{MeV}\)
Total Energy Released:
The total energy released by both reactions is the sum of the energies from each reaction:
\(\Delta E_{\text{total}} = \Delta E_1 + \Delta E_2 = 5.25 + 16.97 = 22.22 \, \text{MeV}\)
Therefore, the energy released during the explosion is 22.22 MeV.
Approach Solution -2
Step 1: Total Reaction - Combining the two reactions, we get:
\(3^{3}Li^{6}\) + \(^{1}H^{2} \rightarrow 2\)\(^{2}\)\(He^{4}\)
Step 2: Calculate Energy Released (Q-value) - Using the mass-energy equivalence \(Q = \Delta m \cdot c^2\), where \(\Delta m\) is the mass defect. - Given:
\(M(^{3}Li^{6}) = 6.01690 \, \text{amu}\)
\(M(^{1}H^{2}) = 2.01471 \, \text{amu}\)
\(M(^{2}He^{4}) = 4.00388 \, \text{amu}\)
\(1 \, \text{amu} = 931.5 \, \text{MeV}\)
Step 3: Calculate Q
\(Q = \left[M(^{3}Li^{6}) + M(^{1}H^{2}) - 2 \times M(^{2}He^{4})\right] \times 931.5 \, \text{MeV}\)
\(Q = \left[6.01690 + 2.01471 - 2 \times 4.00388\right] \times 931.5 \, \text{MeV}\)
\(Q = \left[8.03161 - 8.00776\right] \times 931.5 \, \text{MeV}\)
\(Q = 0.02385 \times 931.5 \, \text{MeV}\)
\(Q = 22.22 \, \text{MeV}\)
So, the correct answer is: 22.22 MeV
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