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Chapter 43: Problem 61

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 \(\frac{198}{80} \mathrm{Hg}_{\cdot} \))

Short Answer

Expert verified
0.958 MeV

Step by step solution

01

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.
02

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:\[ \Delta m = 197.968225 \text{ u} - 197.966752 \text{ u} = 0.001473 \text{ u} \]
03

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).
04

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.
05

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:\[ \text{K.E.}_\text{max} = \Delta E - \text{Photon energy} = 1.37 \text{ MeV} - 0.412 \text{ MeV} = 0.958 \text{ MeV} \]

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Key Concepts

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

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.

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Most popular questions from this chapter

The radioactive nuclide \(^{199} \mathrm{Pt}\) has a half-life of 30.8 minutes. A sample is prepared that has an initial activity of \(7.56 \times 10^{11} \mathrm{Bq}\). (a) How many \(^{199} \mathrm{Pt}\) nuclei are initially present in the sample? (b) How many are present after 30.8 minutes? What is the activity at this time? (c) Repeat part (b) for a time 92.4 minutes after the sample is first prepared.

The United States uses \(1.0 \times 10^{20} \mathrm{J}\) of electrical energy per year. If all this energy came from the fission of \(^{235} \mathrm{U},\) which releases 200 MeV per fission event, (a) how many kilograms of 235 \(\mathrm{U}\) would be used per year and (b) how many kilograms of uranium would have to be mined per year to provide that much \(^{235} \mathrm{U} ?\) (Recall that only 0.70\(\%\) of naturally occurring uranium is \(^{235} \mathrm{U} .\) )

BIO 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 \(^{60} \mathrm{Co}\) is 5000 \(\mathrm{Ci} .\) When its activity falls below \(3500 \mathrm{Ci},\) it is considered too weak a source to use in treatment. You work in the radiology department of a large hospital. One of these \(^{60} \mathrm{Co}\) sources in your inventory was manufactured on October \(6,2004\) . It is now April \(6,2007\) . Is the source still usable? The half-life of \(^{60} \mathrm{Co}\) is 5.271 years.

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)?

The atomic mass of \(_{12}^{25} \mathrm{Mg}\) is 24.985837 \(\mathrm{u},\) and the atomic mass of \(^{25}_{13}\) Al is 24.990429 u. (a) Which of these nuclei will decay into the other? (b) What type of decay will occur? Explain how you determined this. (c) How much energy (in MeV) is released in the decay?
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