Energy input is required to fuse medium-mass nuclei, such as iron or cobalt, into more massive nuclei. Explain why.

Short Answer

Expert verified

There is no energy release because heavier nuclei have lower binding energy than their constituents. So, it requires energy input for the nuclear reaction.

Step by step solution

01

Define nuclei

A focal point around which additional elements are organised or assembled.

02

Explanation

Examine why energy is required for the fusion of medium mass nuclei (such as cobalt or iron). We have extra energy that we call the binding energy when elements fuse or fission because of the mass difference between final nuclei and constituent nuclei, which is defined as,

\(\begin{aligned} BINDING\,\,ENERGY &= \left( {\Delta m} \right){c^2}\\ &= \left( {\left( {Z{m_p} + N{m_n}} \right) - {m_{tot}}} \right){c^2}\end{aligned}\)

As a result, the missing mass in heavier nuclei is proportional to the binding energy in the two medium mass nuclei. One can observe from the book's Figure\({\rm{31}}{\rm{.27}}\)that medium-sized nuclei have the maximum binding energy of all the nuclei. As a result, heavier nuclei fused from medium mass nuclei have lower binding energy than the two nuclei from which they were created. As a result, because the binding energy of a heavier nucleus is lower than the binding energy of two component medium mass nuclei, the energy will not be released. We must inject energy into this nuclear reaction since there is no extra energy in this scenario.

Therefore, as heavier nuclei have lower binding energy than their components, there is no energy release, hence energy input is required for the nuclear reaction to occur.

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Most popular questions from this chapter

(a) Neutron activation of sodium, which is \({\rm{100\% }}\,{\,^{{\rm{23}}}}{\rm{Na}}\), produces\(^{{\rm{24}}}{\rm{Na}}\), which is used in some heart scans, as seen in Table 32.1. The equation for the reaction is \(^{23}Na + n{ \to ^{24}}Na + \gamma \). Find its energy output, given the mass of \(^{{\rm{24}}}{\rm{Na}}\) is \(23.990962\,u\).

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Fallout from nuclear weapons tests in the atmosphere is mainly \({}^{{\rm{90}}}{\rm{Sr}}\) and \({}^{137}Cs\) , which have \(28.6\) - and \(32.2y\) half-lives, respectively. Atmospheric tests were terminated in most countries in 1963, although China only did so in 1980. It has been found that environmental activities of these two isotopes are decreasing faster than their half-lives. Why might this be?

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(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?

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(a) Calculate the energy released in the neutron-induced fission reaction\(n{ + ^{239}}Pu{ \to ^{96}}Sr{ + ^{140}}Ba + 4n\), given \(m{(^{96}}Sr) = 95.921750{\rm{ }}u\)

And

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