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Chapter 32: Problem 32

E Go In one type of fusion reaction a proton fuses with a neutron to form a deuterium nucleus: $$ { }_{1}^{1} \mathrm{H}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{1}^{2} \mathrm{H}+\gamma $$ The masses are \({ }_{1} \mathrm{H}(1.0078 \mathrm{u}),{ }_{0}^{1} \mathrm{n}(1.0087 \mathrm{u}),\) and \({ }_{1}^{2} \mathrm{H}(2.0141 \mathrm{u}) .\) The \(\gamma\) -ray photon is massless. How much energy (in \(\mathrm{MeV}\) ) is released by this reaction?

Short Answer

Expert verified
2.236 MeV is released.

Step by step solution

01

Understand the Reaction

The fusion reaction involves a proton \( ^1_1\text{H} \) and a neutron \( ^1_0\text{n} \) combining to form a deuterium nucleus \( ^2_1\text{H} \) and a gamma photon \( \gamma \). The goal is to calculate the energy released in this reaction.
02

Calculate Initial Mass

The initial mass is the sum of the masses of the proton and neutron. Using the given masses, we have:\[\text{Initial mass} = 1.0078\, \text{u} + 1.0087\, \text{u} = 2.0165\, \text{u}\]
03

Calculate Final Mass

The final mass is the mass of the deuterium nucleus, since the gamma photon is massless. Thus:\[\text{Final mass} = 2.0141\, \text{u}\]
04

Calculate Mass Defect

The mass defect is the difference between the initial and final masses:\[\text{Mass defect} = 2.0165\, \text{u} - 2.0141\, \text{u} = 0.0024\, \text{u}\]
05

Convert Mass Defect to Energy

Use Einstein's equation \( E = mc^2 \) to convert the mass defect to energy. First, convert the mass defect from unified atomic mass units to energy in MeV. The conversion factor is \( 1\, \text{u} = 931.5\, \text{MeV/c}^2 \):\[E = 0.0024\, \text{u} \times 931.5\, \text{MeV/u} = 2.2356\, \text{MeV}\]
06

Conclusion

The energy released by the reaction is approximately \( 2.236\, \text{MeV} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Deuterium
Deuterium is a fascinating form of hydrogen. It is also known as heavy hydrogen because it carries a neutron along with a proton in its nucleus, unlike regular hydrogen, which only has a proton. This heavier form of hydrogen is represented as two diatomic notations:
  • A mass number of 2 (from one proton and one neutron)
  • An atomic number of 1, since there is only one proton in the nucleus.
Because of this unique structure, deuterium plays a critical role in fusion reactions. As it combines with other nuclei, it provides the conditions required for releasing large amounts of energy. Deuterium is abundant enough that it's found naturally in water, though it's rare compared to regular hydrogen.
Mass Defect
The concept of mass defect is crucial in nuclear physics. It represents the small difference between the mass of separated nucleons (such as protons and neutrons) and the mass of the nucleus itself once they are combined. This mass difference plays a significant role because it translates to energy, according to Einstein's famous equation, \(E = mc^2\). In essence, the mass defect is:
  • Initial mass minus final mass after a nuclear reaction.
  • The 'missing' mass, which is converted into energy.
For example, in the fusion reaction discussed, the total initial mass is slightly higher than the mass of the deuterium formed, indicating that some mass has been converted to energy, thus demonstrating the principle of mass-energy equivalence.
Energy Conversion
The conversion of mass to energy is a fundamental principle of nuclear reactions, especially evident in processes like fusion and fission. During these reactions, a portion of the nuclear mass is transformed into energy, manifesting the relationship through Einstein's equation, \(E = mc^2\). Key points regarding energy conversion in nuclear reactions:
  • Matter is not lost but converted into energy.
  • These reactions release tremendous energy due to the high conversion factor (1 u = 931.5 MeV).
As seen in the exercise, a small mass defect results in the release of a significant amount of energy, emphasizing the efficiency and potential of nuclear fusion reactions.
Gamma Photon
Gamma photons are a type of electromagnetic radiation, much like X-rays but even more energetic. When fusion occurs, one common byproduct is the emission of gamma rays, due to:
  • Their ability to carry energy away from the nucleus.
Despite having no mass or charge, gamma photons are crucial in reactions because they distribute energy, helping cool rapidly the fusion products. These photons:
  • Travel at the speed of light.
  • Possess very short wavelengths, making them highly penetrative.
Thus, gamma photons are significant not just in contributing to energy balance in fusion reactions, but also in medical and industrial applications, showcasing the versatile nature of nuclear radiation.

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Most popular questions from this chapter

E Blo Over a full course of treatment, two different tumors are to receive the same absorbed dose of therapeutic radiation. The smaller of the tumors (mass \(=0.12 \mathrm{kg})\) absorbs a total of \(1.7 \mathrm{J}\) of energy. (a) Determine the absorbed dose, in Gy. (b) What is the total energy absorbed by the larger of the tumors (mass \(=0.15 \mathrm{kg}\) )?

E SSM When a \({ }_{92}^{23} \mathrm{U}\) ( \(235.043924 \mathrm{u}\) ) nucleus fissions, about \(200 \mathrm{MeV}\) of energy is released. What is the ratio of this energy to the rest energy of the uranium nucleus?

M GO The water that cools a reactor core enters the reactor at \(216^{\circ} \mathrm{C}\) and leaves at \(287^{\circ} \mathrm{C}\). (The water is pressurized, so it does not turn to steam.) The core is generating \(5.6 \times 10^{9} \mathrm{W}\) of power. Assume that the specific heat capacity of water is \(4420 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) over the temperature range stated above, and find the mass of water that passes through the core each second.

E N-HINT The energy released by each fission within the core of a nuclear reactor is \(2.0 \times 10^{2}\) MeV. The number of fission's occurring each second is \(2.0 \times 10^{19}\). Determine the power (in watts) that the reactor generates.

E ssm Suppose a neutrino is created and has an energy of 35 MeV. (a) Assuming the neutrino, like the photon, has no mass and travels at the speed of light, find the momentum of the neutrino. (b) Determine the de Broglie wavelength of the neutrino.
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