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Chapter 43: Problem 44

Consider the nuclear reaction $$ _{14}^{28} \mathrm{Si}+\gamma \rightarrow_{12}^{24} \mathrm{Mg}+\mathrm{X} $$ 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 \(_{14}^{28}\) Si atom is 27.976927 \( \mathrm{u},\) and the mass of a \(^{24}_{12} \mathrm{Mg}\) atom is 23.985042 \(\mathrm{u}\)

Short Answer

Expert verified
(a) \(Z=2\), \(A=4\) for \(X\). (b) Minimum photon energy is 9.987 MeV.

Step by step solution

01

Identify Conservation Laws

First, we apply the conservation of atomic number and mass number in nuclear reactions. The atomic number is the number of protons, denoted as \(Z\), and the mass number is the number of protons plus neutrons, denoted as \(A\). For the reaction given, the total \(Z\) and \(A\) on the left side (\(^{28}_{14} \text{Si} + \gamma\)) must equal the total \(Z\) and \(A\) on the right side (\(^{24}_{12} \text{Mg} + X\)).
02

Conservation of Atomic Number (Z)

Using conservation of atomic number: \(Z_{\text{Si}} + 0 = Z_{\text{Mg}} + Z_X\). Here, \(Z_{\text{Si}} = 14\) and \(Z_{\text{Mg}} = 12\), so \(14 = 12 + Z_X\). Solving for \(Z_X\), we find \(Z_X = 2\).
03

Conservation of Mass Number (A)

Using conservation of mass number: \(A_{\text{Si}} + 0 = A_{\text{Mg}} + A_X\). Here, \(A_{\text{Si}} = 28\) and \(A_{\text{Mg}} = 24\), so \(28 = 24 + A_X\). Solving for \(A_X\), we find \(A_X = 4\).
04

Determine the Identity of Nuclide X

The identified nuclide \(X\) with \(Z = 2\) and \(A = 4\) is \(^{4}_{2} \text{He}\), which is an alpha particle.
05

Calculate Minimum Energy for Photon

To calculate the minimum energy of the photon (\(\gamma\)), we use mass-energy equivalence. The energy needed equals the mass defect times \(c^2\). Mass defect is given by \[ \Delta m = (m_{\text{X}} + m_{\text{Mg}}) - m_{\text{Si}} \], where \(m_{\text{X}} = 4.002602 \, \text{u}\) (mass of \(^{4}_{2}\text{He}\)), \(m_{\text{Mg}} = 23.985042\, \text{u}\), and \(m_{\text{Si}} = 27.976927 \, \text{u}\). Substituting and solving: \[ \Delta m = (4.002602 + 23.985042) - 27.976927 = 0.010717 \, \text{u}\].
06

Convert Mass Defect to Energy

To convert the mass defect to energy: \[E = \Delta m \cdot c^2\]. Using \(1 \, \text{u} = 931.5 \, \text{MeV}/c^2\), \[ E = 0.010717 \, \text{u} \times 931.5 \, \text{MeV/u} = 9.987 \, \text{MeV}\].

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Key Concepts

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

Conservation of Atomic Number
In nuclear reactions, the conservation of atomic number is a fundamental principle. This means that the total number of protons, denoted by the atomic number \(Z\), must remain unchanged before and after the reaction. For the presented reaction \(_{14}^{28} \mathrm{Si} + \gamma \rightarrow_{12}^{24} \mathrm{Mg} + \mathrm{X}\),\ the atomic number on the left must equal that on the right.

In this scenario:
  • The atomic number of silicon \(^{28}_{14} \text{Si }\) is 14, and a photon \(\gamma\) has an atomic number of 0.
  • The magnesium \(^{24}_{12} \text{Mg}\) on the right side has an atomic number of 12.
  • So, we set up the equation: \(Z_{\text{Si}} + Z_\gamma = Z_{\text{Mg}} + Z_X\).
Solving this, \(14 + 0 = 12 + Z_X\), gives us \(Z_X = 2\). Therefore, the nuclide \(X\) must have an atomic number of 2.
Conservation of Mass Number
The conservation of mass number is another crucial concept in nuclear reactions. The mass number, denoted by \(A\), is the sum of protons and neutrons in a nucleus. In any nuclear reaction, the total mass number on the reactant side must equal the total on the product side.

For our reaction:
  • The mass number of silicon \( _{14}^{28} \text{Si}\) is 28, and the mass number of the photon \(\gamma\) is 0.
  • Magnesium \(^{24}_{12} \text{Mg}\), has a mass number of 24.
  • So, we use the equation: \(A_{\text{Si}} + A_\gamma = A_{\text{Mg}} + A_X\).
This calculation, \(28 + 0 = 24 + A_X\), results in \(A_X = 4\). Hence, the nuclide \(X\) must have a mass number of 4. Together with its atomic number \(Z\), \(_2^4\text{He}\) is identified as the nuclide \(X\).
Mass-Energy Equivalence
Mass-energy equivalence, expressed by Einstein's famous equation \(E = mc^2\), is pivotal in understanding nuclear processes. It relates mass (\(m\)) to energy (\(E\)) and explains how tiny amounts of mass can be converted into significant energy.

In the case of this nuclear reaction:
  • We calculate the mass defect, or the difference between the masses of the products and reactants.
  • Using the mass defect \(\Delta m\), we determine the energy required by the photon to initiate the reaction.
Here, mass defect is found with \(\Delta m = (m_{\text{X}} + m_{\text{Mg}}) - m_{\text{Si}} = 0.010717 \text{ u}\).

The energy equivalent of this mass defect is calculated by converting to energy: \(E = \Delta m \cdot c^2\). With \(1 \text{ u} = 931.5 \text{ MeV}/c^2\), \(E\) computes to approximately 9.987 MeV.
Photon Energy Calculation
To usher a nuclear reaction as shown, the photon must provide energy that at least matches the mass defect calculated earlier. This ensures the reaction can proceed energetically.

The steps to finding this energy include:
  • Identifying which particles we are comparing in terms of mass.
  • Calculating the mass defect \(\Delta m\), which is pivotal in determining the required photon energy.
  • Using the energy-mass conversion factor to convert mass into energy.
We find: \(E = \Delta m \times 931.5\,\text{MeV/u} = 9.987\,\text{MeV}\). This is the minimum photon energy needed for the reaction \(_{14}^{28} \mathrm{Si} + \gamma \rightarrow_{12}^{24} \mathrm{Mg} + \mathrm{X}\) to occur. Photon energy must meet or exceed this value for the reaction to happen.

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Most popular questions from this chapter

(a) Prove that when a particle with mass \(m\) and kinetic energy \(K\) collides with a stationary particle with mass \(M,\) the total kinetic energy \(K_{\mathrm{cm}}\) in the center-of-mass coordinate system (the energy available to cause reactions) is $$K_{\mathrm{cm}}=\frac{M}{M+m} K$$ Assume that the kinetic energies of the particles and nuclei are much lower than their rest energies. (b) If \(K_{\text { th }}\) is the minimum, or threshold, kinetic energy to cause an endoergic reaction to occur in the situation of part (a), show that $$K_{\mathrm{th}}=-\frac{M+m}{M} Q$$

BIO A 67 -kg person accidentally ingests 0.35 Ci of tritium. (a) Assume that the tritium spreads uniformly throughout the body and that each decay leads on the average to the absorption of 5.0 \(\mathrm{keV}\) of energy from the electrons emitted in the decay. The half-life of tritium is \(12.3 \mathrm{y},\) and the RBE of the electrons is \(1.0 .\) Calculate the absorbed dose in rad and the equivalent dose in rem during one week. (b) The \(\beta^{-}\) decay of tritium releases more than 5.0 keV of energy. Why is the average energy absorbed less than the total energy released in the decay?

BIO (a) If a chest x ray delivers 0.25 \(\mathrm{mSv}\) to 5.0 \(\mathrm{kg}\) of tissue, how many total joules of energy does this tissue receive? (b) Natural radiation and cosmic rays deliver about 0.10 \(\mathrm{mSv}\) per year at sea level. Assuming an \(\mathrm{RBE}\) of \(1,\) how many rem and rads is this dose, and how many joules of energy does a 75 -kg person receive in a year? (c) How many chest xays like the one in part (a) would it take to deliver the same total amount of energy to a 75 -kg person as she receives from natural radiation in a year at sea level, as described in part (b)?

In the 1986 disaster at the Chernobyl reactor in the Soviet Union (now Ukraine), about \(\frac{1}{8}\) of the \(^{137} \mathrm{Cs}\) present in the reactor was released. The isotope \(^{137} \mathrm{Cs}\) has a half-life for \(\beta\) decay of 30.07 \(\mathrm{y}\) and decays with the emission of a total of 1.17 \(\mathrm{MeV}\) of energy per decay. Of this, 0.51 \(\mathrm{MeV}\) goes to the emitted electron and the remaining 0.66 \(\mathrm{MeV}\) to a \(\gamma\) ray. The radioactive \(^{137} \mathrm{Cs}\) is absorbed by plants, which are eaten by livestock and humans. How many \(^{137} \mathrm{Cs}\) atoms would need to be present in each kilogram of body tissue if an equivalent dose for one week is 3.5 \(\mathrm{Sv}\) ? Assume that all of the energy from the decay is deposited in that 1.0 \(\mathrm{kg}\) of tissue and that the RBE of the electrons is 1.5.

We Are Stardust. In 1952 spectral lines of the element technetium- 99\(\left(^{99} \mathrm{Tc}\right)\) 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 \(^{99} \mathrm{Tc}\) is \(200,000\) years. (a) For how many half-lives has the \(^{99} \mathrm{Tc}\) been in the red-giant star if its age is 10 billion years? (b) What fraction of the original \(^{99} \mathrm{Tc}\) 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 \(^{99} \mathrm{Tc}\) 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."
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