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

Short Answer

Expert verified
The binding energy per nucleon is approximately 11.49 MeV/nucleon.

Step by step solution

01

Understand the Problem

The goal is to find the binding energy per nucleon for the isotope \( ^{259}_{104} \text{Rf} \). To do this, we will need to calculate the total binding energy of the nucleus and then divide it by the number of nucleons (protons and neutrons) it contains.
02

Calculate the Theoretical Mass of the Nucleus

The theoretical mass of the nucleus is calculated by the sum of the protons and neutrons if they were not bound together. For \( ^{259}_{104} \text{Rf} \), this means:- Protons: 104 \( \times \) mass of a proton (let's use hydrogen atom mass) = 104 \( \times \) 1.007825 u- Neutrons: 155 \( \times \) mass of a neutron = 155 \( \times \) 1.008665 uCalculate:\[104 \times 1.007825 + 155 \times 1.008665\]
03

Calculate the Mass Defect

The mass defect is found by subtracting the actual atomic mass from the theoretical mass calculated:\( \text{Mass defect} = (104 \times 1.007825 + 155 \times 1.008665) - 259.10563 \mathrm{u} \).
04

Convert Mass Defect to Binding Energy

To find the binding energy from the mass defect, use Einstein's equation \( E = \Delta m \times c^2 \), where\( \Delta m \) is the mass defect, and \( c \) is the speed of light. We know 1 amu = 931.5 MeV/c², so:\[\text{Binding energy (MeV)} = \text{Mass defect (u)} \times 931.5 \text{ MeV/u}\]
05

Calculate Binding Energy per Nucleon

Finally, divide the total binding energy by the number of nucleons (259) for Rutherfordium-259:\[\text{Binding energy per nucleon} = \frac{\text{Total binding energy (MeV)}}{259}\]
06

Perform Calculations

Let's plug in the numbers from the previous steps:**Theoretical mass:**\(104 \times 1.007825 + 155 \times 1.008665 = 262.302175\) u**Mass defect:**\(262.302175 - 259.10563 = 3.196545\) u**Binding energy:**\(3.196545 \times 931.5 \approx 2976.37\) MeV**Binding energy per nucleon:**\(\frac{2976.37}{259} \approx 11.49\) MeV/nucleon.

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Key Concepts

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

Mass Defect
In nuclear physics, the concept of mass defect is crucial to understanding nuclear binding energy. When nucleons (protons and neutrons) combine to form a nucleus, the mass of the nucleus is less than the sum of the individual masses of the protons and neutrons. This difference in mass is known as the mass defect.

The mass defect occurs because when nucleons are brought together to form a nucleus, energy is released. This energy is what holds the nucleus together, and it must be supplied to break the nucleus apart. According to mass-energy equivalence, a principle established by Einstein, the lost mass is converted into energy, thus binding the nucleons together.

To determine the mass defect, one calculated the theoretical mass of a nucleus by summing the masses of its constituent protons and neutrons. This calculated mass is then compared to the nucleus's actual measured atomic mass. The difference between these two is the mass defect, which can be used to determine the binding energy of the nucleus.
Atomic Mass
Atomic mass is the weighted average mass of an element's isotopes based on their natural abundance. For a single atom, the atomic mass is almost equivalent to the mass number (the sum of protons and neutrons) but slightly less due to a phenomenon known as mass defect.

Atomic mass is expressed in atomic mass units (amu), where 1 amu is defined as one-twelfth of the mass of a carbon-12 atom. Each element's atomic mass reflected on the periodic table accounts for the mixture of isotopes and their relative masses. In the context of binding energy calculations, the atomic mass is integral in determining the mass defect and further calculating nuclear binding energy.

Understanding atomic mass helps us comprehend atomic structure and nuclear energy calculations. It provides a basis for predicting how a nucleus will behave under certain physical conditions and how energy is conserved and expended during nuclear reactions.
Nucleons
Nucleons are the particles that make up the atomic nucleus, comprising protons and neutrons. These particles are held together by the strong nuclear force, one of the fundamental forces in nature.

- **Protons**: Positively charged particles that determine the element's identity. The number of protons is called the atomic number. - **Neutrons**: Neutrally charged particles that add mass and stability to the nucleus. Neutrons help in balancing the electromagnetic force between protons, preventing the nucleus from breaking apart due to electrostatic repulsion.

The sum of protons and neutrons in a nucleus is the mass number, which is closely associated with the atomic mass. In nuclear calculations, such as determining the mass defect or binding energy, the number of nucleons is significant because it influences the amount of energy that binds them together. Knowing the number of nucleons in an isotope helps define the isotope's properties and its stability.
Einstein's Equation
Einstein's equation, expressed as \( E = mc^2 \), is a fundamental principle in physics, demonstrating the relationship between mass and energy. This equation highlights the concept of mass-energy equivalence, indicating that mass can be converted into energy and vice versa.

In the context of nuclear binding energy, Einstein's equation allows us to calculate the binding energy of a nucleus from its mass defect. The equation states that the energy (\( E \)) is equal to the mass (\( m \)) multiplied by the speed of light (\( c \)) squared. This relationship shows that even a small amount of mass can be converted into a large amount of energy.

For example, when calculating the binding energy of a nucleus, the mass defect is converted into energy using the conversion factor where 1 amu corresponds to 931.5 MeV/c². By transforming mass losses into energy during nuclear formation, Einstein's equation underscores how nuclear processes can result in significant energy changes, key in both physics and practical applications like nuclear power generation.

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Most popular questions from this chapter

The radionuclide \({ }^{11} \mathrm{C}\) decays according to $$ { }^{11} \mathrm{C} \rightarrow{ }^{11} \mathrm{~B}+\mathrm{e}^{+}+\nu, \quad T_{1 / 2}=20.3 \min $$ 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.)

What is the activity of a \(10 \mathrm{ng}\) sample of \({ }^{92} \mathrm{Kr}\), which has a halflife of \(1.84 \mathrm{~s}\) ?

The isotope \({ }^{40} \mathrm{~K}\) can decay to either \({ }^{40} \mathrm{Ca}\) or \({ }^{40} \mathrm{Ar} ;\) assume both decays have a half-life of \(1.26 \times 10^{9} \mathrm{y} .\) The ratio of the Ca produced to the Ar produced is \(8.54 / 1=8.54\). A sample originally had only \({ }^{40} \mathrm{~K}\). It now has equal amounts of \({ }^{40} \mathrm{~K}\) and \({ }^{40} \mathrm{Ar}\); that is, the ratio of \(\mathrm{K}\) to Ar is \(1 / 1=1\). How old is the sample? (Hint: Work this like other radioactive-dating problems, except that this decay has two products.)

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 \(250 \mathrm{Ci}\) ?

Consider an initially pure \(3.4 \mathrm{~g}\) sample of \({ }^{67} \mathrm{Ga}\), an isotope that has a half-life of \(78 \mathrm{~h}\). (a) What is its initial decay rate? (b) What is its decay rate \(48 \mathrm{~h}\) later?

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