Problem 57 - Fundamentals of Physics

Question

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.)

Step-by-Step Solution

Answer

(a) Derived expression for Q is correct. (b) Calculated Q is 0.961 MeV, which is similar to the given maximum positron energy of 0.960 MeV.

Step 1: Understanding Disintegration Energy Q

The disintegration energy \( Q \) of a nuclear reaction is the difference in the initial and final mass-energy of the system. It accounts for the energy released when a radionuclide decays into another nuclide. This energy is shared by the decay products: the daughter nuclide, the emitted positron, and the neutrino.

Step 2: Derive the Expression for Q

To find the disintegration energy, use the equation:\[ Q = \left(m_{\text{initial}} - m_{\text{final}}\right) c^2 \]Since the nuclear masses are not directly observed, we must use atomic masses, taking into account the electrons.The reaction is:\[ {}^{11}\mathrm{C} \rightarrow {}^{11}\mathrm{B} + \mathrm{e}^{+} + u \]The disintegration energy \( Q \) is the mass difference of the original and final atoms, converted to energy:\[ Q = (m_{\text{C}} - m_{\text{B}} - 2m_{\text{e}})c^2 \]The term \(2m_{\text{e}}\) accounts for the positron mass emitted and binding differences in electrons.

Step 3: Calculate Q Using Given Mass Values

Given masses:- \( m_{\text{C}} = 11.011434 \) u- \( m_{\text{B}} = 11.009305 \) u- \( m_{\text{e}} = 0.0005486 \) uSubstitute these into the equation for \( Q \):\[ Q = \left(11.011434 - 11.009305 - 2 \times 0.0005486 \right) c^2 \]\[ Q = (0.0010318) c^2 \]Convert atomic mass unit energy equivalence: 1 u \( \approx 931.5 \) MeV:\[ Q \approx 0.0010318 \times 931.5 \] MeV \[ Q \approx 0.961 \] MeV.

Step 4: Compare Q with Maximum Positron Energy

The maximum energy of emitted positrons is given as 0.960 MeV. The calculated disintegration energy \( Q \approx 0.961 \) MeV. Positrons will have almost all of this energy, except that carried away by the neutrino. The small difference due to the neutrino energy is expected.

Key Concepts

Disintegration Energy

Disintegration energy is an important concept in nuclear physics. It refers to the energy released during a nuclear decay process. When a radionuclide undergoes decay, the masses of the initial and final states are different. This mass difference is converted into energy, which is what we term as the disintegration energy. In the given exercise, the disintegration energy \( Q \) for the decay of \( {}^{11} \mathrm{C} \) is calculated using the equation:

• \[ Q = (m_{\mathrm{C}} - m_{\mathrm{B}} - 2m_{\mathrm{e}})c^2 \]

Here, \( m_{\mathrm{C}} \) is the mass of carbon-11, \( m_{\mathrm{B}} \) is the mass of boron-11, and \( 2m_{\mathrm{e}} \) accounts for the mass of the emitted positron and the associated electrons. The energy \( Q \) is shared among all decay products, like positrons and neutrinos, allowing us to see why these particles carry energy away from the process.

Atomic Mass

Atomic mass plays a crucial role in nuclear decay calculations. It represents the mass of an atom, with its protons, neutrons, and the surrounding electron cloud. The exercise uses the atomic masses of \( {}^{11}\mathrm{C} \) and \( {}^{11}\mathrm{B} \):

• \( m_{\mathrm{C}} = 11.011434 \; \text{u} \)
• \( m_{\mathrm{B}} = 11.009305 \; \text{u} \)

The atomic mass includes the masses of the electrons, which must be considered to account correctly for mass changes during nuclear decay. Proper application of atomic masses ensures precise calculations of the disintegration energy, making them fundamental to understanding nuclear reactions and the conservation of energy principle inherent in these processes. By converting atomic masses to energy units (\( 1 \; \text{u} \approx 931.5 \; \text{MeV} \)), we can calculate the corresponding energy transformations.

Nuclear Decay

Nuclear decay refers to the process by which an unstable atomic nucleus loses energy by radiation. In the exercise, the decay of \( {}^{11}\mathrm{C} \) results in the formation of \( {}^{11}\mathrm{B} \), a positron, and a neutrino:

• \[ {}^{11} \mathrm{C} \rightarrow {}^{11} \mathrm{B} + \mathrm{e}^{+} + u \]

This process happens because \( {}^{11}\mathrm{C} \) is radioactive and seeks a more stable state.
Nuclear decay types include alpha, beta, and gamma decay. This specific decay process involves positron emission, which is a type of beta decay. Understanding nuclear decay processes is crucial because these reactions power various processes, from the sun’s energy output to the operation of nuclear reactors. Knowing how energy is conserved and converted in decay events helps us harness these reactions for energy and medical applications.

Positron Emission

Positron emission, an intriguing aspect of nuclear decay, involves the emission of a positron (\( \mathrm{e}^{+} \)), which is a positively charged particle akin to the electron but with opposite charge. In the decay of \( {}^{11}\mathrm{C} \), a positron is emitted along with a neutrino. The positron's maximum energy is given as 0.960 MeV.

• This energy corresponds to the excess energy not carried away by the neutrino.
• Positron emission is a way for a proton-rich nucleus to become more stable by converting a proton into a neutron.

Because positrons are antimatter counterparts of electrons, they annihilate upon meeting electrons, producing gamma rays. This phenomenon is used in medical imaging techniques, such as positron emission tomography (PET scans), to observe inside the human body. By examining positron emissions during nuclear decay, we gain insights into both fundamental particle interactions and practical applications in technology and medicine.