How does the lithium deuteride in the thermonuclear bomb shown in Figure \({\rm{32}}{\rm{.33}}\) supply tritium (\({}^3H\)) as well as deuterium (\({}^2H\))?

Short Answer

Expert verified

Deuterium is derived from the remaining mixture of lithium deuteride, whereas tritium is obtained via the fission process of lithium deuteride.

Step by step solution

01

Define thermonuclear weapon

A thermonuclear weapon, often known as a fusion weapon or hydrogen bomb (H bomb), is a type of nuclear weapon developed in the second generation.

02

Explanation

In this study, we look at how lithium deuteride supplies tritium and deuterium in the thermonuclear bomb's reaction,

\({}^6Li{}^2H + n \to {}^4He + {}^3H\)

We can observe that this reaction occurs in the presence of neutrons, indicating that nuclear fission is the cause. Since tritium is being produced (the fission product). There is a chance of fusion reaction now that we have tritium and deuterium in a lithium combination, which might provide a massive quantity of free energy. As a result,

\({}_1^2H + {}_1^3H \to {}_2^4He + {}_0^1n + 17.59\,MeV\)

As shown, tritium is created from lithium deuteride and nuclear fission (which is initiated by the neutron). With surplus neutron and roughly\({\rm{17 MeV}}\)of free energy, residual deuterium from the lithium mixture combines with tritium and fuses to helium.

Helium is only an inert by-product in this reaction. It is the free energy of roughly\({\rm{17 MeV}}\)that poses a serious threat because\(\gamma \)rays and kinetic energy can be emitted.

Therefore, tritium is obtained from the lithium deuteride fission process, whereas deuterium is obtained from the lithium deuteride residual mixture.

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Most popular questions from this chapter

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Find the amount of energy given to the \(^4He\) nucleus and to the \(\gamma \) ray in the reaction\(n{ + ^3}He{ \to ^4}He + \gamma \), using the conservation of momentum principle and taking the reactants to be initially at rest. This should confirm the contention that most of the energy goes to the \(\gamma \) ray.

The power output of the Sun is \(4 \times {10^{26}}{\rm{ }}W\).

(a) If \(90\% \) of this is supplied by the proton-proton cycle, how many protons are consumed per second?

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How many kilograms of water are needed to obtain the \(198.8{\rm{ }}mol\) of deuterium, assuming that deuterium is \(0.01500\% \) (by number) of natural hydrogen?

(a) Estimate the years that the deuterium fuel in the oceans could supply the energy needs of the world. Assume world energy consumption to be ten times that of the United States which is \(8 \times {10^9}J/y\) and that the deuterium in the oceans could be converted to energy with an efficiency of \(32\% \). You must estimate or look up the amount of water in the oceans and take the deuterium content to be \(0.015\% \) of natural hydrogen to find the mass of deuterium available. Note that approximate energy yield of deuterium is \(3.37 \times {10^{14}}J/kg\).

(b) Comment on how much time this is by any human measure. (It is not an unreasonable result, only an impressive one.)

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