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Give reasons justifying the contention made in the text that energy from the fusion reaction \({}^2H + {}^2H \to {}^4He + \gamma \) is relatively difficult to capture and utilize.

Short Answer

Expert verified

Experimentally, it is quite difficult to properly trap and absorb \({\rm{\gamma }}\) rays.

Step by step solution

01

Define energy

Energy is a measurable attribute that may be transferred from one thing to another in order for it to do work.

02

Explanation

Let’s investigate why the following nuclear reaction is difficult to capture in this problem,

\({}^2H + {}^2H \to {}^4He + \gamma \)

The helium core and\(\gamma \)rays, which are formed when the nucleus core shifts from one excited state to another excited state or the ground state, are visible as a result of this nuclear process. When we look at the electromagnetic spectrum, we can observe that\(\gamma \)rays have extraordinarily high frequencies, as well as incredibly high energy, if we use the Einstein connection\(E = hf\). Since we know that\(c = \lambda f\), this has a very tiny wavelength since it is of high frequency.

The\(\gamma \)rays are exceedingly destructive and might cause harm if they reach the Earth's surface. However, due to their incredibly tiny wavelength, it would be exceedingly difficult to collect and identify these\(\gamma \)rays even in the nuclear reactor where this reaction would occur. Producing a detector that can properly absorb these rays is extremely difficult, and it also raises the issue of potential health risks during the surgery if it occurs.

Therefore, experiments in capturing and absorbing \(\gamma \) rays have proven to be quite difficult.

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Most popular questions from this chapter

(a) Calculate the energy released in the neutron-induced fission (similar to the spontaneous fission in Example\(32.3\)) \(n{ + ^{238}}U{ \to ^{96}}Sr{ + ^{140}}Xe + 3n\), given \(m{(^{96}}Sr) = 95.921750{\rm{ }}u\) and \(m{(^{140}}Xe) = 139.92164{\rm{ }}u\).

(b) This result is about \(6{\rm{ }}MeV\) greater than the result for spontaneous fission. Why?

(c) Confirm that the total number of nucleons and total charge are conserved in this reaction.

The naturally occurring radioactive isotope \(^{{\rm{232}}}{\rm{Th}}\) does not make good fission fuel, because it has an even number of neutrons; however, it can be bred into a suitable fuel (much as \(^{{\rm{238}}}{\rm{U}}\) is bred into\(^{239}P\)).

(a) What are Z and N for\(^{{\rm{232}}}{\rm{Th}}\)?

(b) Write the reaction equation for neutron captured by \(^{{\rm{232}}}{\rm{Th}}\) and identify the nuclide \(^AX\)produced in\(n{ + ^{232}}Th{ \to ^A}X + \gamma \).

(c) The product nucleus \({\beta ^ - }\)decays, as does its daughter. Write the decay equations for each, and identify the final nucleus.

(d) Confirm that the final nucleus has an odd number of neutrons, making it a better fission fuel.

(e) Look up the half-life of the final nucleus to see if it lives long enough to be a useful fuel.

Verify by listing the number of nucleons, total charge, and electron family number before and after the cycle that these quantities are conserved in the overall proton-proton cycle in \(2{e^ - } + {4^1}H{ \to ^4}He + 2{\nu _e} + 6\gamma \).

Two fusion reactions mentioned in the text are

\(n{ + ^3}He{ \to ^4}He + \gamma \)

and

\(n{ + ^1}H{ \to ^2}H + \gamma \).

Both reactions release energy, but the second also creates more fuel. Confirm that the energies produced in the reactions are \(20.58\) and\(2.22{\rm{ }}MeV\), respectively. Comment on which product nuclide is most tightly bound, \(^4He\) or\(^2H\).

(a) What temperature gas would have atoms moving fast enough to bring two \(^{\rm{3}}{\rm{He}}\) nuclei into contact? Note that, because both are moving, the average kinetic energy only needs to be half the electric potential energy of these doubly charged nuclei when just in contact with one another.

(b) Does this high temperature imply practical difficulties for doing this in controlled fusion?

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