/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 What is the binding energy per n... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What is the binding energy per nucleon of the rutherfordium isotope \({ }_{104}^{259} \mathrm{Rf}\) ? Here are some atomic masses and the neutron mass. $$\begin{array}{lr}\frac{259}{104} \mathrm{Rf} & 259.10563 \mathrm{u} \\\\\mathrm{n} & 1.008665 \mathrm{u}\end{array} { }^{1} \mathrm{H} \quad 1.007825 \mathrm{u}$$

Short Answer

Expert verified
The binding energy per nucleon of ${}_{104}^{259} \mathrm{Rf}$ is approximately 9.51 MeV/nucleon.

Step by step solution

01

Calculate the Mass Defect

The mass defect is the difference between the sum of the masses of the individual nucleons (protons and neutrons) and the actual mass of the nucleus. The number of protons in Rutherfordium (Rf) is 104, so the number of neutrons is \(259 - 104 = 155\). Calculate the expected mass of nucleons: \((104 \times 1.007825) + (155 \times 1.008665)\). \[ \text{Mass of nucleons} = 104 \times 1.007825 + 155 \times 1.008665 \approx 261.749775 \text{ u} \]The mass defect \(\Delta m\) is:\[ \Delta m = 261.749775 - 259.10563 = 2.644145 \text{ u} \]
02

Convert Mass Defect to Energy

We use Einstein’s mass-energy equivalence principle \(E = \Delta m \cdot c^2\) where \(c\) is the speed of light \(\approx 3 \times 10^8 \text{ m/s}\). The conversion factor from atomic mass units to energy is \(1 \text{ u} = 931.5 \text{ MeV}\). Therefore:\[ E = 2.644145 \times 931.5 = 2463.2651175 \text{ MeV} \]
03

Calculate the Binding Energy per Nucleon

To find the binding energy per nucleon, divide the total binding energy by the total number of nucleons.\[ \text{Binding energy per nucleon} = \frac{2463.2651175}{259} \approx 9.5103 \text{ MeV/nucleon} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Mass Defect
In nuclear physics, mass defect is a crucial concept. It refers to the difference between the total mass of individual nucleons when they are free and the actual mass of a nucleus when these nucleons are bound together. Specifically, once protons and neutrons are assembled into a nucleus, the total mass is reduced compared to when they are separate. This is because energy is required to separate these nucleons, reflecting Einstein’s famous equation that relates mass and energy: \( E = mc^2 \).
The mass defect is calculated by subtracting the nucleus' actual mass from the sum of the masses of its constituent protons and neutrons. For Rutherfordium \( ({}_{104}^{259} \text{Rf}) \), with its calculated nucleon mass approximately 261.75 atomic mass units (u), and its actual mass being 259.10563 u, the mass defect is approximately 2.644 u. This difference signifies the binding energy that holds the nucleus together.
Mass-Energy Equivalence
The concept of mass-energy equivalence is central to understanding how the mass defect translates into binding energy. According to Einstein's mass-energy relation \( E = mc^2 \), mass and energy can be converted into one another. In the context of nuclear binding energy, the mass defect reflects the energy required to bind the nucleus.
To convert the mass defect to energy, we use the conversion factor where 1 atomic mass unit (u) equals approximately 931.5 Mega-electronvolts (MeV). Thus, a mass defect of 2.644 u for Rutherfordium (Rf) transforms to approximately 2463.27 MeV. This energy amount reveals how much energy is necessary to break the nucleus into its individual nucleons. Understanding this connection between mass and energy helps us gauge the stability of an atomic nucleus.
Atomic Masses
Atomic masses are pivotal in nuclear physics calculations. They are typically measured in atomic mass units (u), with each isotope having a specific atomic mass. For instance, the masses of an individual proton (hydrogen atom \( {}^1\text{H} \)) and neutron used in calculations are 1.007825 u and 1.008665 u respectively.
Atomic masses play an essential role in determining the mass defect. For the isotope \({}_{104}^{259} \text{Rf}\), the number of protons is 104, and the nucleus also contains 155 neutrons. By multiplying the number of each nucleon by their respective atomic masses and summing these quantities, we deduce the expected combined nucleon mass. This process thus allows us to compute the mass defect and, by extension, the binding energy, leveraging the relationship between mass and energy.
Nucleon Calculations
Nucleon calculations are fundamental to understanding nuclear binding energies. A nucleon refers to either a proton or neutron in an atomic nucleus. The calculation involves determining the number of protons and neutrons. For the isotope \( {}_{104}^{259} \text{Rf} \), there are 104 protons and 155 neutrons in each nucleus.
The mass of these nucleons is calculated by multiplying their quantities with their respective atomic masses (protons 1.007825 u and neutrons 1.008665 u). These include:
  • The total proton mass: \( 104 \times 1.007825 \)
  • The total neutron mass: \( 155 \times 1.008665 \)
The process of combining these calculated masses enables us to estimate the expected mass of the entire nucleus. This expected mass, when compared to the actual nuclear mass provided by atomic measurements, lets us compute the mass defect. These nucleon calculations are a key step to understand both the atomic scale interactions and the strengths of the forces within the nucleus.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The half-life of a radioactive isotope is 140 d. How many days would it take for the decay rate of a sample of this isotope to fall to one-fourth of its initial value?

The nuclide \({ }^{198} \mathrm{Au},\) with a half-life of \(2.70 \mathrm{~d},\) is used in cancer therapy. What mass of this nuclide is required to produce an activity of \(250 \mathrm{Ci} ?\)

Go An \(\alpha\) particle ('He nucleus) is to be taken apart in the following steps. Give the energy (work) required for each step: (a) remove a proton, (b) remove a neutron, and (c) separate the remaining proton and neutron. For an \(\alpha\) particle, what are (d) the total binding energy and (e) the binding energy per nucleon? (f) Does either match an answer to (a), (b), or (c)? Here are some atomic masses and the neutron mass. $$ \begin{array}{llll} { }^{4} \mathrm{He} & 4.00260 \mathrm{u} & { }^{2} \mathrm{H} & 2.01410 \mathrm{u} \\ { }^{3} \mathrm{H} & 3.01605 \mathrm{u} & { }^{1} \mathrm{H} & 1.00783 \mathrm{u} \\ \mathrm{n} & 1.00867 \mathrm{u} & & \end{array} $$

The radioactive nuclide \({ }^{99}\) Tc can be injected into a patient's bloodstream in order to monitor the blood flow, measure the blood volume, or find a tumor, among other goals. The nuclide is produced in a hospital by a "cow" containing \({ }^{99} \mathrm{Mo},\) a radioactive nuclide that decays to \({ }^{99} \mathrm{Tc}\) with a half-life of \(67 \mathrm{~h}\). Once a day, the cow is "milked" for its \({ }^{99} \mathrm{Tc},\) which is produced in an excited state by the \({ }^{99} \mathrm{Mo} ;\) the \({ }^{99} \mathrm{Tc}\) de- excites to its lowest energy state by emitting a gamma-ray photon, which is recorded by detectors placed around the patient. The de-excitation has a half- life of \(6.0 \mathrm{~h}\). (a) By what process does \({ }^{99}\) Mo decay to \({ }^{99} \mathrm{Tc} ?\) (b) If a patient is injected with an \(8.2 \times 10^{7}\) Bq sample of \({ }^{99} \mathrm{Tc}\), how many gamma-ray photons are initially produced within the patient each second? (c) If the emission rate of gamma-ray photons from a small tumor that has collected \({ }^{99} \mathrm{Tc}\) is 38 per second at a certain time, how many excited-state \({ }^{99} \mathrm{Tc}\) are located in the tumor at that time?

The radionuclide \({ }^{11} \mathrm{C}\) decays according to $${ }^{11} \mathrm{C} \rightarrow{ }^{11} \mathrm{~B}+\mathrm{e}^{+}+\nu, \quad T_{1 / 2}=20.3 \mathrm{~min}$$ The maximum energy of the emitted positrons is \(0.960 \mathrm{MeV}\). (a) Show that the disintegration energy \(Q\) for this process is given by $$Q=\left(m_{\mathrm{C}}-m_{\mathrm{B}}-2 m_{\mathrm{e}}\right) c^{2}$$ where \(m_{\mathrm{C}}\) and \(m_{\mathrm{B}}\) are the atomic masses of \({ }^{11} \mathrm{C}\) and \({ }^{11} \mathrm{~B}\), respectively, and \(m_{e}\) is the mass of a positron. (b) Given the mass values \(m_{\mathrm{C}}=11.011434 \mathrm{u}, m_{\mathrm{B}}=11.009305 \mathrm{u},\) and \(m_{\mathrm{e}}=\) \(0.0005486 \mathrm{u},\) calculate \(Q\) and compare it with the maximum energy of the emitted positron given above. (Hint: Let \(\mathbf{m}_{\mathrm{C}}\) and \(\mathbf{m}_{\mathrm{B}}\) be the nuclear masses and then add in enough electrons to use the atomic masses.)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.