Question

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 \( 1 \, \text{amu} = 931.5 \, \text{MeV} \)]

• A. 28.12 MeV
• B. 12.64 MeV
• C. 16.48 MeV
• D. 22.22 MeV

Answer 1

The Correct Option is D

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