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Chapter 42: Problem 44

What is the binding energy per nucleon of the americium isotope \({ }_{95}^{244} \mathrm{Am} ?\) Here are some atomic masses and the neutron mass. \(\begin{array}{lrrr}{ }_{95}^{244} \mathrm{Am} & 244.064279 \mathrm{u} & { }^{1} \mathrm{H} & 1.007825 \mathrm{u} \\ \mathrm{n} & 1.008665 \mathrm{u} & & \end{array}\)

Short Answer

Expert verified
The binding energy per nucleon of \(\text{ }^{244}_{95}\text{Am} \) is approximately 7.52 MeV/nucleon.

Step by step solution

01

Identify the Number of Protons and Neutrons

The isotope of americium given is \( \text{ }^{244}_{95}\text{Am} \), which means it has 95 protons and a mass number of 244. To find the number of neutrons, subtract the number of protons from the mass number: \( 244 - 95 = 149 \). So, there are 149 neutrons.
02

Find the Total Mass of Free Protons and Neutrons

To calculate the mass of 95 protons, use the mass of hydrogen (which is approximate to a proton) as \( 1.007825 \text{ u} \). The total mass for the protons is \( 95 \times 1.007825 = 95.743375 \text{ u} \). For 149 neutrons, multiply by the given neutron mass \( 1.008665 \text{ u} \): \( 149 \times 1.008665 = 150.289385 \text{ u} \). The sum of these is the total mass of free protons and neutrons: \( 95.743375 + 150.289385 = 246.03276 \text{ u} \).
03

Calculate the Mass Defect

The mass defect is the difference between the total mass of the free nucleons and the actual atomic mass of \( \text{ }^{244}_{95}\text{Am} \). Use the given atomic mass of Americium: \( 244.064279 \text{ u} \). The mass defect \( \Delta m \) is \( 246.03276 - 244.064279 = 1.968481 \text{ u} \).
04

Calculate the Binding Energy

Convert the mass defect in atomic mass units to energy using Einstein's equation: \( E = \Delta m \cdot c^2 \). The conversion factor is \( 1 \text{ u} = 931.5 \text{ MeV/c}^2 \). Therefore, the binding energy \( E \) is \( 1.968481 \times 931.5 = 1833.9152875 \text{ MeV} \).
05

Determine the Binding Energy per Nucleon

To find the binding energy per nucleon, divide the total binding energy by the total number of nucleons (mass number \( A = 244 \)). So, \( \frac{1833.9152875}{244} = 7.517288 \text{ MeV/nucleon} \).

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Key Concepts

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

Mass Defect
The concept of mass defect is central to understanding nuclear binding energy. Mass defect refers to the difference between the sum of the masses of protons and neutrons separately and the actual mass of the nucleus. This happens because energy is released when a nucleus is formed, leading to a loss of mass according to the conservation of energy. This missing mass
  • Accounts for the energy needed to hold the nucleus together, known as the binding energy.
  • Is computed based on the formula: \[ \Delta m = \text{Sum of separate nucleon masses} - \text{Actual nucleus mass} \]
  • Is often measured in atomic mass units (u).
Understanding mass defect is crucial because it signifies how much mass is converted into binding energy during the formation of a nucleus. In our example with americium, the observed mass defect shows how the sum of individual nucleon masses differs from the measured atomic mass.
Nuclear Physics
Nuclear physics is the branch of physics focused on the components and behavior of atomic nuclei. It delves into phenomena such as radioactivity, nuclear reactions, and nuclear structure. Key aspects of nuclear physics include:
  • Protons and neutrons, the particles within a nucleus, are collectively known as nucleons.
  • The strong nuclear force, a fundamental force, binds nucleons together, overcoming the repulsion between protons due to their positive charge.
  • Binding energy, which is the energy required to de-assemble a nucleus into its separate protons and neutrons, showcases the stability of a nucleus. Higher binding energy indicates greater stability.
In exercises like the one on americium, calculating binding energy per nucleon offers insights into the energy dynamics within the nucleus. It reveals how tightly the nucleus' constituents are bound, reflecting the stability and energy potential of nuclear matter.
Einstein's Equation
Einstein's equation, famously known as \( E=mc^2 \), plays a fundamental role in connecting mass and energy. This equation:
  • Shows that energy and mass are interchangeable. A small mass can be converted into a significant amount of energy.
  • Is crucial for understanding nuclear processes where tiny mass changes translate to considerable energy release or absorption.
  • Provides the conversion factor needed for calculating binding energy from mass defects. In nuclear reactions, the mass defect (in atomic mass units) is converted to energy (in MeV) using this principle, \[ E = \Delta m \times c^2 \quad \text{where} \quad 1 \text{ u} = 931.5 \text{ MeV/c}^2 \].
The application of Einstein's equation helps students appreciate how nuclear fusion and fission release large amounts of energy, making it an integral part of understanding nuclear physics and energy phenomena.

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Most popular questions from this chapter

The isotope \({ }^{238} \mathrm{U}\) decays to \({ }^{206} \mathrm{~Pb}\) with a half-life of \(4.47 \times 10^{9} \mathrm{y}\). Although the decay occurs in many individual steps, the first step has by far the longest half-life; therefore, one can often consider the decay to go directly to lead. That is, \({ }^{238} \mathrm{U} \rightarrow{ }^{206} \mathrm{~Pb}+\) various decay products. A rock is found to contain \(4.20 \mathrm{mg}\) of \({ }^{238} \mathrm{U}\) and \(2.135 \mathrm{mg}\) of \({ }^{206} \mathrm{~Pb}\). Assume that the rock contained no lead at formation, so all the lead now present arose from the decay of uranium. How many atoms of (a) \({ }^{238} \mathrm{U}\) and (b) \({ }^{206} \mathrm{~Pb}\) does the rock now contain? (c) How many atoms of \({ }^{238} \mathrm{U}\) did the rock contain at formation? (d) What is the age of the rock?

Plutonium isotope \({ }^{239} \mathrm{Pu}\) decays by alpha decay with a half- life of \(24100 \mathrm{y}\). How many milligrams of helium are produced by an initially pure \(10.0 \mathrm{~g}\) sample of \({ }^{239} \mathrm{Pu}\) at the end of \(20000 \mathrm{y}\) ? is emitted by a nucleus, along with a neutrino. The emitted particles share the available disintegration energy. The electrons and positrons emitted in beta decay have a continuous spectrum of energies from near zero up to a limit \(K_{\max }\left(=Q=-\Delta m c^{2}\right)\). Radioactive Dating Naturally occurring radioactive nuclides provide a means for estimating the dates of historic and prehistoric events. For example, the ages of organic materials can often be found by measuring their \({ }^{14} \mathrm{C}\) content; rock samples can be dated using the radioactive isotope \({ }^{40} \mathrm{~K}\). Radiation Dosage Three units are used to describe exposure to ionizing radiation. The becquerel ( \(1 \mathrm{~Bq}=1\) decay per second) measures the activity of a source. The amount of energy actually absorbed is measured in grays, with 1 Gy corresponding to \(1 \mathrm{~J} / \mathrm{kg}\). The estimated biological effect of the absorbed energy is measured in sieverts; a dose equivalent of 1 Sv causes the same biological effect regardless of the radiation type by which it was acquired. Nuclear Models The collective model of nuclear structure assumes that nucleons collide constantly with one another and that relatively long-lived compound nuclei are formed when a projectile is captured. The formation and eventual decay of a compound nucleus are totally independent events. The independent particle model of nuclear structure assumes that each nucleon moves, essentially without collisions, in a quantized state within the nucleus. The model predicts nucleon levels and magic nucleon numbers \((2,8,20,28,50,82\), and 126) associated with closed shells of nucleons; nuclides with any of these numbers of neutrons or protons are particularly stable. The combined model, in which extra nucleons occupy quantized states outside a central core of closed shells, is highly successful in predicting many nuclear properties.

A \(6.00 \mathrm{~g}\) charcoal sample from an ancient fire pit has a \({ }^{14} \mathrm{C}\) activity of \(63.0\) disintegrations/min. A living tree has a \({ }^{14} \mathrm{C}\) activity of \(15.3\) disintegrations/min per \(1.00 \mathrm{~g}\). The half-life of \({ }^{14} \mathrm{C}\) is \(5730 \mathrm{y}\) How old is the charcoal sample?

An organic sample of mass \(4.00 \mathrm{~kg}\) absorbs \(2.00 \mathrm{~mJ}\) via slow neutron radiation \((\mathrm{RBE}=5)\). What is the dose equivalent \((\mathrm{mSv})\) ?

A radioactive sample intended for irradiation of a hospital patient is prepared at a nearby laboratory. The sample has a halflife of \(83.61 \mathrm{~h}\). What should its initial activity be if its activity is to be \(7.4 \times 10^{8} \mathrm{~Bq}\) when it is used to irradiate the patient \(24 \mathrm{~h}\) later?
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