Problem 61 - University Physics with Modern Physics

Question

Gold, \(_{79}^{198} \mathrm{Au}\), undergoes \(\beta^{-}\) decay to an excited state of \(^{198}_{80} \mathrm{Hg}\). If the excited state decays by emission of a photon with energy 0.412 MeV, what is the maximum kinetic energy of the electron emitted in the decay? This maximum occurs when the antineutrino has negligible energy. (The recoil energy of the \(^{198}_{80} \mathrm{Hg}\) nucleus can be ignored. The masses of the neutral atoms in their ground states are 197.968225 u for \(^{198}_{80} \mathrm {Au}\) and 197.966752 u for \(\frac{198}{80} \mathrm{Hg}_{\cdot} \))

Step-by-Step Solution

Answer

0.958 MeV

Step 1: Understand the Process of Beta Decay

In beta decay, a neutron in the nucleus of an atom decays into a proton, an electron (also known as a beta particle), and an antineutrino. For this problem, gold \[ _{79}^{198} \mathrm{Au} \] decays to an excited state of mercury \( _{80}^{198} \mathrm{Hg} \) by emitting a beta particle and an antineutrino.

Step 2: Calculate the Mass Difference

We start by finding the mass difference \( \Delta m \) between the parent and the daughter nucleus. The mass of the parent \( _{79}^{198} \mathrm{Au} \) is 197.968225 u and that of the daughter \( _{80}^{198} \mathrm{Hg} \) is 197.966752 u. The mass difference is given by:\[ \Delta m = 197.968225 \text{ u} - 197.966752 \text{ u} = 0.001473 \text{ u} \]

Step 3: Convert Mass Difference to Energy

Convert the mass difference \( \Delta m \) to energy using Einstein's mass-energy equivalence principle, \( E = \Delta m \cdot c^2 \), where \( c \) is the speed of light \( 3 \times 10^8 \text{ m/s} \). The energy in MeV is given by:\[ \Delta E = 0.001473 \text{ u} \times 931.5 \text{ MeV/u} = 1.37 \text{ MeV} \] (since 1 u = 931.5 MeV).

Step 4: Determine Energy of Emitted Photon

The problem states that the excited state of \( ^{198}_{80} \mathrm{Hg} \) emits a photon with an energy of 0.412 MeV when transitioning to its ground state. This energy is not available to the beta particle.

Step 5: Calculate Maximum Kinetic Energy of Electron

The energy available for the electron and the antineutrino is the total energy difference \( \Delta E \) minus the photon energy. Since the maximum kinetic energy of the electron occurs when the antineutrino has negligible energy, this is given by:\[ \text{K.E.}_\text{max} = \Delta E - \text{Photon energy} = 1.37 \text{ MeV} - 0.412 \text{ MeV} = 0.958 \text{ MeV} \]

Key Concepts

Kinetic Energy

Kinetic energy is a form of energy that a particle possesses due to its motion. When we talk about kinetic energy in the context of beta decay, we're focusing on the energy of the emitted electron, also called a beta particle. In beta-minus decay, a neutron in the nucleus transforms into a proton while emitting a beta particle and an antineutrino.

• The emitted electron carries away kinetic energy which is part of the energy released during the decay.
• The maximum kinetic energy of this electron is defined when the antineutrino absorbs minimal energy.

In the actual exercise, the maximum kinetic energy is calculated by subtracting the energy of the emitted photon (0.412 MeV) from the total energy difference (1.37 MeV) of the nuclear reaction. This calculation shows how kinetic energy distribution among decay products is crucial for determining the kinetics of the decay process.

Mass-Energy Equivalence

Mass-energy equivalence is a fundamental principle in physics, famously encapsulated by Einstein's equation: \[ E = mc^2 \]This equation tells us that mass can be converted into energy and vice versa.
In a nuclear reaction such as beta decay, a small difference in mass between the parent and daughter nuclei results in a large amount of energy being released.

• In the given problem, this energy difference arises from a mass deficit of 0.001473 atomic mass units (u), converted into energy using the factor 931.5 MeV/u.
• As a result, the total energy released is 1.37 MeV.

Understanding this conversion is key to analyzing nuclear processes, making it possible to predict the energy carried away by decay products like electrons, antineutrinos, and photons.

Photon Emission

Photon emission often occurs when an atom transitions from a higher energy state to a lower energy state. In the context of beta decay, this process is crucial in balancing the energy of the nuclear reaction.
When gold ( _{79}^{198} ext{Au} ) undergoes beta decay to an excited state of mercury ( _{80}^{198} ext{Hg} ), the transition of mercury from its excited to ground state releases a photon with 0.412 MeV energy.

• This photon emission reduces the energy that can be imparted to other particles, such as the electron.
• By accounting for this photon energy, we determine the maximum kinetic energy that the beta particle can inherit.

Photon emission is a pivotal aspect of nuclear decay, influencing the energy dynamics of particles generated in the process.