Problem 44 - University Physics with Modern Physics

Question

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 \(\mathrm{u}\)

Step-by-Step Solution

Answer

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

Step 1: 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 (\(^{24}_{12} \text{Mg} + X\)).

Step 2: 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 \(Z_X = 2\).

Step 3: 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 \(A_X = 4\).

Step 4: 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.

Step 5: 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: \[ \Delta m = (4.002602 + 23.985042) - 27.976927 = 0.010717 \, \text{u}\].

Step 6: Convert Mass Defect to Energy

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

Key Concepts

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.