The explosive in a Hydrogen bomb is a mixture of
_1mathrmH^2${}_{1}\mathrm{H}^{2}$,
_1mathrmH^3${}_{1}\mathrm{H}^{3}$ and
_3mathrmLi^6${}_{3}\mathrm{Li}^{6}$ in some condensed form. The chain reaction is given by:
beginarrayl _ 3 mathrm L i ^ 6 + _ 0 mathrm n ^ 1 rightarrow _ 2 mathrm H e ^ 4 + _ 1 mathrm H ^ 3 \\ _ 1 mathrm H ^ 2 + _ 1 mathrm H ^ 3 rightarrow _ 2 mathrm H e ^ 4 + _ 0 mathrm n ^ 1 endarray$$\begin{array}{l} { } _ { 3 } \mathrm { L i } ^ { 6 } + { } _ { 0 } \mathrm { n } ^ { 1 } \rightarrow { } _ { 2 } \mathrm { H e } ^ { 4 } + { } _ { 1 } \mathrm { H } ^ { 3 } \\ { } _ { 1 } \mathrm { H } ^ { 2 } + { } _ { 1 } \mathrm { H } ^ { 3 } \rightarrow { } _ { 2 } \mathrm { H e } ^ { 4 } + { } _ { 0 } \mathrm { n } ^ { 1 } \end{array}$$
During the explosion the energy released is approximately:
[Given:
mathbfM(mathrmLi) = 6.01690 mathrm~amu$\mathbf{M}(\mathrm{Li}) = 6.01690 \mathrm{~amu}$,
mathbfM(_1mathbfH^2) = 2.01471 mathrm~amu$\mathbf{M}({}_1\mathbf{H}^2) = 2.01471 \mathrm{~amu}$,
mathbfM(_2mathbfHe^4) = 4.00388 mathrm~amu$\mathbf{M}({}_2\mathbf{H}e^4) = 4.00388 \mathrm{~amu}$, and
1 mathrm~amu = 931.5 mathrm~MeV$1 \mathrm{~amu} = 931.5 \mathrm{~MeV}$]
Solution
### Related Formula
The Q-value or energy released (Q$Q$) during a nuclear reaction sequence is determined from mass defect (Delta m$\Delta m$):
Q = Delta m times 931.5 mathrm~MeV$$Q = \Delta m \times 931.5 \mathrm{~MeV}$$
### Core Logic
Adding the two equations together to obtain the single combined net nuclear reaction:
_3mathrmLi^6 + _0mathrmn^1 + _1mathrmH^2 + _1mathrmH^3 rightarrow 2left(_2mathrmHe^4right) + _1mathrmH^3 + _0mathrmn^1$${}_{3}\mathrm{Li}^{6} + {}_{0}\mathrm{n}^{1} + {}_{1}\mathrm{H}^{2} + {}_{1}\mathrm{H}^{3} \rightarrow 2\left({}_{2}\mathrm{He}^{4}\right) + {}_{1}\mathrm{H}^{3} + {}_{0}\mathrm{n}^{1}$$
Cancelling intermediate species appearing on both sides yields:
_3mathrmLi^6 + _1mathrmH^2 rightarrow 2left(_2mathrmHe^4right)$${}_{3}\mathrm{Li}^{6} + {}_{1}\mathrm{H}^{2} \rightarrow 2\left({}_{2}\mathrm{He}^{4}\right)$$
### Step 1: Calculate Mass Defect
The mass defect Delta m$\Delta m$ of this net process is:
Delta m = M(mathrmLi) + M(_1mathrmH^2) - 2 M(_2mathrmHe^4)$$\Delta m = M(\mathrm{Li}) + M({}_{1}\mathrm{H}^{2}) - 2 M({}_{2}\mathrm{He}^{4})$$
Substituting the given mass profiles:
Delta m = 6.01690 + 2.01471 - 2(4.00388)$$\Delta m = 6.01690 + 2.01471 - 2(4.00388)$$
Delta m = 8.03161 - 8.00776 = 0.02385 mathrm~amu$$\Delta m = 8.03161 - 8.00776 = 0.02385 \mathrm{~amu}$$
### Step 2: Compute Energy Released
Converting mass defect into MeV value:
Q = 0.02385 times 931.5 mathrm~MeV approx 22.216 mathrm~MeV$$Q = 0.02385 \times 931.5 \mathrm{~MeV} \approx 22.216 \mathrm{~MeV}$$
Rounding off gives approximately 22.22 mathrm~MeV$22.22 \mathrm{~MeV}$.
### Pattern Recognition
When chain equations share intermediate steps (like neutron consumption/generation or tritium tracking), add the algebraic steps together to deduce the overall net target process. This cuts out unnecessary individual constituent mass balances.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
