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

Answer: (D)

22.22 MeV

Answer 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