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Chapter 18: Problem 44

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 1.0827 MeV.

Step by step solution

01

Determine the number of protons and neutrons in the nucleus

First, we need to determine the number of protons and neutrons in the \(^{56}\mathrm{Fe}\) nucleus. Since it's an iron atom, it has 26 protons. To find the number of neutrons, subtract the number of protons from the mass number (56): \( 56 - 26 = 30 \) neutrons.
02

Calculate the total mass of individual nucleons

Next, we need to find the total mass of the individual protons and neutrons. The atomic mass unit (u) is approximately equal to the mass of a proton and a neutron, so: Total mass of protons = number of protons × mass of one proton Total mass of protons = 26 × 1u = 26u Total mass of neutrons = number of neutrons × mass of one neutron Total mass of neutrons = 30 × 1u = 30u Total mass of individual nucleons = Total mass of protons + Total mass of neutrons Total mass of individual nucleons = 26u + 30u = 56u
03

Find the mass defect

Now we can find the mass defect, which is the difference between the mass of the nucleus and the total mass of its individual protons and neutrons: Mass defect = Total mass of individual nucleons - Mass of nucleus Mass defect = 56u - 55.9349u = 0.0651u
04

Convert the mass defect into energy

Using Einstein's mass-energy equivalence formula, we can convert the mass defect into energy. The formula is: \(E = mc^2\) Here, E is the energy, m is the mass defect, and c is the speed of light in a vacuum. Since 1u is equal to 931.5 MeV/c², we can convert the mass defect into MeV and then calculate the energy: Mass defect in MeV = 0.0651u × 931.5 MeV/c²/u = 60.6128 MeV
05

Calculate the binding energy per nucleon

Finally, to find the binding energy per nucleon, simply divide the total binding energy by the number of nucleons: Binding energy per nucleon = Total binding energy / Number of nucleons Binding energy per nucleon = 60.6128 MeV / 56 = 1.0827 MeV So the binding energy per nucleon for \(^{56}\mathrm{Fe}\) is approximately 1.0827 MeV.

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Most popular questions from this chapter

A \(0.10-\mathrm{cm}^{3}\) sample of a solution containing a radioactive nuclide \(\left(5.0 \times 10^{3}\) counts per minute per milliliter) is injected \right. into a rat. Several minutes later \(1.0 \mathrm{cm}^{3}\) blood is removed. The blood shows 48 counts per minute of radioactivity. Calculate the volume of blood in the rat. What assumptions must be made in performing this calculation?

The mass ratios of \(^{40} \mathrm{Ar}\) to \(^{40} \mathrm{K}\) also can be used to date geologic materials. Potassium-40 decays by two processes: $$\begin{aligned} &_{19}^{40} \mathrm{K}+_{-1}^{0} \mathrm{e} \longrightarrow_{18}^{40} \mathrm{Ar}(10.7 \%) \quad t_{1 / 2}=1.27 \times 10^{9} \text { years }\\\ &_{19}^{40} \mathrm{K} \longrightarrow_{20}^{40} \mathrm{Ca}+_{-1}^{0} \mathrm{e}(89.3 \%) \end{aligned}$$a. Why are \(^{40} \mathrm{Ar} /^{40} \mathrm{K}\) ratios used to date materials rather than \(^{40} \mathrm{Ca} /^{40} \mathrm{K}\) ratios? b. What assumptions must be made using this technique? c. A sedimentary rock has an \(^{40} \mathrm{Ar} /^{40} \mathrm{K}\) ratio of 0.95. Calculate the age of the rock. d. How will the measured age of a rock compare to the actual age if some \(^{40}\) Ar escaped from the sample?

Using the kinetic molecular theory (see Section 8.6), calculate the root mean square velocity and the average kinetic energy of \(_{1}^{2} \mathrm{H}\) nuclei at a temperature of \(4 \times 10^{7} \mathrm{K}\). (See Exercise 50 for the appropriate mass values.)

Calculate the binding energy per nucleon for \(\frac{2}{1} \mathrm{H}\) and \(^{3}_{1}\) \(\mathrm{H}\). The atomic masses are \(\frac{2}{1} \mathrm{H}, 2.01410 \mathrm{u} ;\) and \(\frac{3}{1} \mathrm{H}, 3.01605 \mathrm{u}\)

Uranium- 235 undergoes many different fission reactions. For one such reaction, when \(^{235} \mathrm{U}\) is struck with a neutron, \(^{144} \mathrm{Ce}\) and \(^{90}\) Sr are produced along with some neutrons and electrons. How many neutrons and \(\beta\) -particles are produced in this fission reaction?
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