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Chapter 43: Problem 46

Consider the nuclear reaction \(^{28}_{14}Si\) + \(\gamma\) \(\rightarrow\) \(^{24}_{12}Mg\) + X where X is a nuclide. (a) What are Z and A for the nuclide X? (b) Ignoring the effects of recoil, what minimum energy must the photon have for this reaction to occur? The mass of a \(^{28}_{14}Si\) atom is 27.976927 u, and the mass of a \(^{24}_{12}Mg\) atom is 23.985042 u.

Short Answer

Expert verified
(a) Z = 2, A = 4. (b) Photon energy = 9.986 MeV.

Step by step solution

01

Understanding the Problem

The given nuclear reaction involves a gamma photon causing a reaction between silicon-28 and magnesium-24, producing another nuclide 'X'. We need to determine the atomic number (Z) and mass number (A) of X and calculate the minimum energy of the photon needed for this reaction.
02

Calculate Atomic Number Z of Nuclide X

In a nuclear reaction, the sum of atomic numbers on both sides must be equal. The initial reactants have Z values of 14 for Silicon and 0 for the photon. On the products side, Magnesium has Z = 12, so X must satisfy:\[ Z_{Si} + Z_{\gamma} = Z_{Mg} + Z_{X} \]\[ 14 + 0 = 12 + Z_{X} \]\[ Z_{X} = 2 \]Therefore, the atomic number Z of X is 2.
03

Calculate Mass Number A of Nuclide X

For mass numbers, the principle of conservation of nucleons is also followed; the sum of mass numbers on both sides of the equation must be equal. Start with:\[ A_{Si} + A_{\gamma} = A_{Mg} + A_{X} \]\[ 28 + 0 = 24 + A_{X} \]\[ A_{X} = 4 \]Thus, the mass number A of X is 4.
04

Calculate the Energy Requirement

The minimum energy of the photon can be calculated using the mass-energy equivalence principle. The energy needed corresponds to the mass defect between the reactants and the products:\[ \Delta m = (m_{Si} - m_{Mg} - m_{X}) \times 931.5 \text{ MeV/u} \]where 931.5 MeV/u is the energy equivalence per atomic mass unit.Assuming \(m_{X} \approx 4.0026 \text{ u}\) (similar to Helium):\[ \Delta m = 27.976927\, \text{u} - (23.985042\, \text{u} + 4.0026 \text{ u})\approx -0.010715 \text{ u}\]So, the energy required for the photon is approximately \[ (-0.010715 \text{ u}) \times 931.5 \text{ MeV/u} \approx -9.986 \text{ MeV} \]. Since energy cannot be negative, the photon must have at least 9.986 MeV to cause the reaction.

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Key Concepts

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

Mass-Energy Equivalence
In nuclear physics, the concept of mass-energy equivalence is essential for understanding how matter and energy interconvert. This principle, famously represented by Einstein's equation \( E=mc^2 \), indicates that a small amount of matter can be converted into a large amount of energy. In nuclear reactions, such as the one involving silicon and magnesium, mass does not remain constant but transforms into energy.
  • The mass of the reactants and products is crucial in calculating the energy changes during a reaction.
  • The energy equivalence of 1 atomic mass unit (u) is approximately 931.5 MeV.
  • When considering energy changes, any mass defect (loss or gain) results in the release or absorption of energy.
For the nuclear reaction in the exercise, the mass defect is converted into the energy required for the reaction to occur. This demonstrates the critical role mass-energy equivalence plays in understanding nuclear processes.
Atomic Number
The atomic number, denoted as \( Z \), represents the number of protons in an atom's nucleus. It determines the element's identity and its position on the periodic table. In nuclear reactions, the conservation of electric charge ensures that the sum of atomic numbers remains constant before and after the reaction.
  • In the given reaction, Silicon \( (^{28}_{14}Si) \) has an atomic number of 14.
  • A gamma photon \( (\gamma) \) contributes 0 to the atomic number as it carries no charge.
  • Magnesium \( (^{24}_{12}Mg) \) has an atomic number of 12, which affects the final product.
To find \( Z \) for nuclide X, the equation \( Z_{Si} + Z_{\gamma} = Z_{Mg} + Z_{X} \) gives \( Z_{X} = 2 \). This indicates nuclide X is Helium, as Helium has an atomic number of 2.
Mass Number
The mass number, denoted \( A \), indicates the total number of protons and neutrons in an atom's nucleus. It plays a vital role in understanding nuclear stability and reactions. Unlike the atomic number, which signifies charge, mass number gives the nucleon count involved in reactions.
  • Silicon in the reaction has a mass number of 28, which contributes to the initial state.
  • The gamma photon \((\gamma)\) carries no nucleons, affecting the mass balance minimally.
  • Magnesium has a mass number of 24, leaving nucleons to be allocated to nuclide X.
Using the relationship \( A_{Si} + A_{\gamma} = A_{Mg} + A_{X} \), we find \( A_{X} = 4 \). This confirms X is Helium, as Helium has a distinct mass number of 4.
Conservation of Nucleons
In nuclear reactions, the conservation of nucleons is a pivotal concept. This principle suggests that the sum of protons and neutrons (nucleons) remains the same across both sides of the reaction equation. This conservation ensures matter stability during the transformations that occur in nuclear changes.
  • The initial nucleon count in silicon and the gamma photon sets a base for comparison.
  • On the product side, magnesium and nuclide X complete the conservation equation.
  • Conserving nucleons ensures no loss in mass, only energy exchange.
For the reaction discussed, maintaining the nucleon total ensures we accurately identify properties of any resulting nuclide. Conservation laws like this make predictions in nuclear physics reliable, as seen in determining nuclide X's characteristics in the reaction.

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Most popular questions from this chapter

The unstable isotope \(^4$$^0\)K is used for dating rock samples. Its half-life is 1.28 \(\times\) 10\(^9\) y. (a) How many decays occur per second in a sample containing 1.63 \(\times\) 10\(^-$$^6\) g of \(^4$$^0\)K? (b) What is the activity of the sample in curies?

\(\textbf{We Are Stardust.}\) In 1952 spectral lines of the element technetium-99 (\(^9$$^9\)Tc) were discovered in a red giant star. Red giants are very old stars, often around 10 billion years old, and near the end of their lives. Technetium has no stable isotopes, and the half-life of \(^9$$^9\)Tc is 200,000 years. (a) For how many halflives has the \(^9$$^9\)Tc been in the red giant star if its age is 10 billion years? (b) What fraction of the original \(^9$$^9\)Tc would be left at the end of that time? This discovery was extremely important because it provided convincing evidence for the theory (now essentially known to be true) that most of the atoms heavier than hydrogen and helium were made inside of stars by thermonuclear fusion and other nuclear processes. If the \(^9$$^9\)Tc had been part of the star since it was born, the amount remaining after 10 billion years would have been so minute that it would not have been detectable. This knowledge is what led the late astronomer Carl Sagan to proclaim that "we are stardust".

\(^2$$^3$$^8\)U decays spontaneously by \(\alpha\) emission to \(^2$$^3$$^4\)Th. Calculate (a) the total energy released by this process and (b) the recoil velocity of the \(^2$$^3$$^4\)Th nucleus. The atomic masses are 238.050788 u for \(^2$$^3$$^8\)U and 234.043601 u for \(^2$$^3$$^4\)Th.

The United States uses about 1.4 \(\times\) 10\(^1$$^9\) J of electrical energy per year. If all this energy came from the fission of \(^2$$^3$$^5\)U, which releases 200 MeV per fission event, (a) how many kilograms of \(^2$$^3$$^5\)U would be used per year, and (b) how many kilograms of uranium would have to be mined per year to provide that much \(^2$$^3$$^5\)U? (Recall that only 0.70% of naturally occurring uranium is \(^2$$^3$$^5\)U.)

Radioactive isotopes used in cancer therapy have a "shelf-life," like pharmaceuticals used in chemotherapy. Just after it has been manufactured in a nuclear reactor, the activity of a sample of \(^6$$^0\)Co is 5000 Ci. When its activity falls below 3500 Ci, it is considered too weak a source to use in treatment. You work in the radiology department of a large hospital. One of these \(^6$$^0\)Co sources in your inventory was manufactured on October 6, 2011. It is now April 6, 2014. Is the source still usable? The half-life of \(^6$$^0\)Co is 5.271 years.
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