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Chapter 21: Problem 93

Modern designs of atomic bombs contain, in addition to uranium or plutonium, small amounts of tritium and deuterium to boost the power of explosion. What is the role of tritium and deuterium in these bombs?

Short Answer

Expert verified
Tritium and deuterium are used in atomic bombs to boost the power of the explosion. When the fission bomb is detonated, the intense heat and pressure cause these elements to undergo nuclear fusion, which releases a significant amount of energy thereby intensifying the overall destructive power of the bomb.

Step by step solution

01

Overview of Nuclear Fusion

Tritium and deuterium are isotopes involved in nuclear fusion, a process where atomic nuclei combine to form a heavier nucleus with the release of energy. This energy is considerably higher compared to that released in a typical atomic bomb, which is based on nuclear fission (splitting of large atomic nuclei).
02

Role of Tritium and Deuterium in Atomic Bomb

The tritium and deuterium are placed in the bomb's core. When the fission bomb is detonated, the intense heat and pressure cause tritium and deuterium to undergo nuclear fusion. The fusion of tritium and deuterium produces helium and a neutron, accompanied by the release of a massive amount of energy.
03

Resulting Effect

The energy released from this fusion reaction significantly increases the overall destructive power of the bomb. Thus, tritium and deuterium serve the role of a 'booster' for the atomic bomb, leading to a far more powerful explosion than what could be achieved by fission alone.

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Most popular questions from this chapter

Nuclear waste disposal is one of the major concerns of the nuclear industry. In choosing a safe and stable environment to store nuclear wastes, consideration must be given to the heat released during nuclear decay. As an example, consider the \(\beta\) decay of \({ }^{90} \mathrm{Sr}\) \((89.907738 \mathrm{amu})\) $$ { }_{38}^{90} \mathrm{Sr} \longrightarrow{ }_{39}^{90} \mathrm{Y}+{ }_{-1}^{0} \beta \quad t_{\frac{1}{2}}=28.1 \mathrm{yr} $$ The \({ }^{90} \mathrm{Y}(89.907152 \mathrm{amu})\) further decays as follows: $$ { }_{39}^{90} \mathrm{Y} \longrightarrow{ }_{40}^{90} \mathrm{Zr}+{ }_{-1}^{0} \beta \quad t_{\frac{1}{2}}=64 \mathrm{~h} $$ Zirconium-90 (89.904703 amu) is a stable isotope. (a) Use the mass defect to calculate the energy released (in joules) in each of the preceding two decays. (The mass of the electron is \(5.4857 \times\) \(10^{-4}\) amu. ( b) Starting with 1 mole of \({ }^{90}\) Sr, calculate the number of moles of \(9^{9}\) Sr that will decay in a year. (c) Calculate the amount of heat released (in kilojoules) corresponding to the number of moles of \({ }^{90} \mathrm{Sr}\) decayed to \({ }^{90} \mathrm{Zr}\) in \((\mathrm{b})\)

For each pair of isotopes listed, predict which one is less stable: (a) \({ }_{3}^{6} \mathrm{Li}\) or \({ }_{3}^{9} \mathrm{Li},\) (b) \({ }_{11}^{23} \mathrm{Na}\) or \({ }_{11}^{25} \mathrm{Na},\) (c) \({ }_{20}^{48} \mathrm{Ca}\) or \({ }_{21}^{48} \mathrm{Sc}\).

After the Chernobyl accident, people living close to the nuclear reactor site were urged to take large amounts of potassium iodide as a safety precaution. What is the chemical basis for this action?

In the thorium decay series, thorium- 232 loses a total of \(6 \alpha\) particles and \(4 \beta\) particles in a 10 -stage process. What is the final isotope produced?

A long-cherished dream of alchemists was to produce gold from cheaper and more abundant elements. This dream was finally realized when \({ }_{80}^{198} \mathrm{Hg}\) was converted into gold by neutron bombardment. Write a balanced equation for this reaction.
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