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Chapter 19: Problem 50

The most stable nucleus in terms of binding energy per nucleon is \(^{56}\mathrm{Fe}\) . If the atomic mass of \(^{56}\mathrm{Fe}\) is $55.9349 \mathrm{u},\( calculate the binding energy per nucleon for \)^{56} \mathrm{Fe} .$

Short Answer

Expert verified
The binding energy per nucleon for \(^{56}\mathrm{Fe}\) is approximately 8.56 MeV.

Step by step solution

01

Identify the number of protons and neutrons

Since the isotope is \(^{56}\mathrm{Fe}\), it has 56 nucleons in total. Iron (Fe) has an atomic number of 26, which means it has 26 protons. To find the number of neutrons, we subtract the number of protons from the total number of nucleons: \(56 - 26 = 30\) neutrons.
02

Calculate the mass defect

The mass defect is the difference between the mass of the individual protons and neutrons and the actual mass of \(^{56}\mathrm{Fe}\). The atomic mass of a proton is approximately 1.00728 u and the atomic mass of a neutron is approximately 1.00867 u. We can calculate the mass defect using the following equation: \[Mass\,Defect = (26 \times 1.00728\,u + 30 \times 1.00867\,u) - 55.9349\,u\] \[Mass\,Defect = (26.18928\,u + 30.2601\,u) - 55.9349\,u\] \[Mass\,Defect = 56.44938\,u - 55.9349\,u\] \[Mass\,Defect = 0.51448\,u\]
03

Convert mass defect to energy

Now we will use Einstein's equation to convert the mass defect to energy. We need to multiply the mass defect by the atomic mass unit constant (\(c^2 = 931.5\,\frac{MeV}{c^2 u}\)) to convert the mass defect from atomic mass units (u) to energy in mega electron volts (MeV). \[Total\,Binding \,Energy = 0.51448\,u \times 931.5\,\frac{MeV}{c^2 u}\] \[Total\,Binding \,Energy = 479.17\,MeV\]
04

Calculate binding energy per nucleon

Finally, we will divide the total binding energy by the number of nucleons (56) to find the binding energy per nucleon. \[Binding\,Energy\,Per\,Nucleon = \frac{479.17\,MeV}{56}\] \[Binding\,Energy\,Per\,Nucleon = 8.56\,MeV\] Thus, the binding energy per nucleon for \(^{56}\mathrm{Fe}\) is approximately 8.56 MeV.

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Most popular questions from this chapter

Calculate the binding energy in J/nucleon for carbon-12 (atomic mass \(=12.0000\) u) and uranium-235 (atomic mass \(=\) 235.0439 u). The atomic mass of \(_{1}^{1} \mathrm{H}\) is 1.00782 \(\mathrm{u}\) and the mass of a neutron is 1.00866 u. The most stable nucleus known is \(^{56}\) Fe $(\text { see Exercise } 50)\( . Would the binding energy per nucleon for \)^{56} \mathrm{Fe}$ be larger or smaller than that of \(^{12} \mathrm{C}\) or \(^{235} \mathrm{U}\) ? Explain.

Do radiotracers generally have long or short half-lives? Explain.

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In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium, Sg, in honor of Glenn T. Seaborg, discoverer of the transuranium elements. a. \(^{263}\) Sg was produced by the bombardment of \(^{249} \mathrm{Cf}\) with a beam of \(^{18} \mathrm{O}\) nuclei. Complete and balance an equation for this reaction. b. \(^{263}\) g decays by \(\alpha\) emission. What is the other product resulting from the \(\alpha\) decay of \(^{263} \mathrm{Sg}\) ?

What is the rate of decay from 1.00 mol of radioactive nuclides having the following half-lives: \(12,000\) years? 12 hours? 12 seconds?
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