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Chapter 43: Problem 15

43.15. The \(\alpha\) decay of \(^{238} \mathrm{U}\) is accompanied by a \(\gamma\) ray of measured wavelength 0.0248 \(\mathrm{nm}\) . This decay is due to a transition of the nucleus between two energy levels. What is the difference in energy (in MeV) between these two levels?

Short Answer

Expert verified
The energy difference is approximately 5.00 MeV.

Step by step solution

01

Convert Wavelength to Energy

To find the energy difference associated with the gamma ray, use the relationship between energy, wavelength, and Planck's constant given by the formula: \[ E = \frac{hc}{\lambda} \]where:- \( h = 6.626 \times 10^{-34} \) J路s (Planck's constant),- \( c = 3.00 \times 10^8 \) m/s (speed of light),- \( \lambda = 0.0248 \) nm (wavelength of the gamma ray, converted to meters it is \( 0.0248 \times 10^{-9} \) m).Substitute these values into the formula to find the energy \( E \).
02

Perform the Calculation

Now perform the calculations using the formula: \[ E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{0.0248 \times 10^{-9}} \]This gives:\[ E = \frac{1.9878 \times 10^{-25}}{0.0248 \times 10^{-9}} \approx 8.010 \times 10^{-15} \text{ Joules} \]
03

Convert Energy to MeV

Convert the energy from joules to MeV. The conversion factor is:\[ 1 \text{ Joule} = 6.242 \times 10^{12} \text{ MeV} \]Multiply the energy in Joules by this conversion factor:\[ E \approx 8.010 \times 10^{-15} \times 6.242 \times 10^{12} \approx 4.999 \text{ MeV} \]
04

Conclusion

The energy difference between the two nuclear levels, as indicated by the gamma ray, is approximately 5.00 MeV.

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Key Concepts

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

Alpha Decay
Alpha decay is a type of radioactive decay where an unstable nucleus emits an alpha particle, which consists of 2 protons and 2 neutrons. This emission results in the reduction of the atomic number by 2 and the mass number by 4. For example, in the alpha decay of uranium-238
  • The uranium nucleus loses an alpha particle to become thorium-234.
  • This transformation involves a release of energy due to the mass difference before and after decay.
  • The newly formed element may be left in an excited state, which can result in additional emissions such as gamma rays.
Understanding alpha decay is crucial in nuclear physics as it lies at the heart of nuclear reactions and affects nuclear stability.
Gamma Ray Spectroscopy
Gamma ray spectroscopy is a technique used to study the energy levels within a nucleus. It involves measuring the energies of gamma rays emitted as a nucleus transitions from a higher to a lower energy state. This method provides invaluable insights into nuclear structure and the processes occurring within it.
  • Gamma rays are highly energetic electromagnetic waves released during nuclear transitions.
  • Each gamma ray energy is unique to a specific transition, making them ideal probes for nuclear energy levels.
  • The precise measurement of gamma ray wavelengths allows for accurate determination of energy differences within a nucleus.
This tool is essential for identifying isotopes and understanding nuclear processes like those involved in the decay of uranium-238.
Energy Level Transition
Energy level transitions occur when a nucleus changes from one energy state to another. This change often involves the emission or absorption of energy in the form of radiation, such as gamma rays.
  • Nuclear transitions are quantized, meaning the energy released or absorbed is specific to the energy difference between two levels.
  • In the uranium-238 example, the transition results in the emission of a gamma ray indicating the energy difference.
  • Such transitions provide crucial information on the internal arrangement of nucleons within a nucleus.
By studying these transitions, scientists gain a deeper understanding of nuclear forces and stability mechanisms.
Planck's Constant
Planck's Constant is a fundamental constant in quantum mechanics that relates the energy of a photon to its frequency. It is denoted as \( h \) and has a value of \( 6.626 \times 10^{-34} \text{ J路s} \).
  • In the context of gamma rays, it helps determine the energy associated with a specific wavelength.
  • The relation is given by the equation \( E = \frac{hc}{\lambda} \), where \( E \) is energy, \( c \) is the speed of light, and \( \lambda \) is wavelength.
  • This fundamental relation is key to understanding electromagnetic radiation behavior in various scenarios.
Its importance extends beyond nuclear physics, impacting fields like quantum electrodynamics and the study of atomic phenomena.
Energy Conversion
Energy conversion in nuclear physics typically involves translating energy from one form to another, such as from joules to electronvolts (eV) or mega-electronvolts (MeV), which are more practical for measuring nuclear energies.
  • For practical use, nuclear energy differences are often expressed in eV or MeV due to the small scale of energy transactions.
  • 1 eV is the amount of energy gained by an electron when it moves through a potential difference of one volt.
  • The conversion from joules involves the factor \( 1 \text{ Joule} = 6.242 \times 10^{12} \text{ MeV} \).
This conversion is crucial for interpreting results from experiments and comparing them across different studies in nuclear and particle physics.

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Most popular questions from this chapter

43.12. The most common isotope of copper is \(\frac{63}{29} \mathrm{Cu}\) . The measured mass of the neutral atom is 62.929601 u. (a) From the measured mass, determine the mass defect, and use it to find the total binding energy and the binding energy per nucleon. (b) Calculare the binding energy from Eq. ( 43.11\()\) . (Why is the fifth term zero? Compare to the result you obtained in part (a). What is the percent difference? What do you conclude about the accuracy of Eq. \((43.11) ?\)

43.68. A \(70.0-\mathrm{kg}\) person experiences a whole-body exposure to \(\alpha\) radiation with energy 4.77 \(\mathrm{MeV}\) . A total of \(6.25 \times 10^{12} \alpha\) particles are absorbed. (a) What is the absorbed dose in rad? (b) What is the equivalent dose in rem?(c) If the source is 0.0320 \(\mathrm{g}\) of 26 \(\mathrm{Ra}\) (half- life 1600 \(\mathrm{y}\) ) somewhere in the body, what is the activity of this source? (d) If all the alpha particles produced are absorbed, what time is required for this dose to be delivered?

43.39. A \(50-\mathrm{kg}\) person accidentally ingests 0.35 \(\mathrm{Ci}\) 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 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?

43.28. The ratio of \(^{14} \mathrm{C}\) to \(^{12} \mathrm{C}\) in living matter is measured to be \(^{14} \mathrm{C} /^{2} \mathrm{C}=1.3 \times 10^{-12}\) at the present time. A \(12.0-\mathrm{g}\) sample of carbon produces 180 decays/min due to the small amount of \(^{14} \mathrm{C}\) in it. From this information, calculate the half-life of "C.

43.67 Measurements indicate that 27.83\(\%\) of all rubidium atoms currently on the earth are the radioactive \(^{87} \mathrm{Rb}\) isotope. The rest are the stable \(^{85} \mathrm{Rb}\) isotope. The half-life of \(^{87} \mathrm{Rb}\) is 4.75 \(\times 10^{10} \mathrm{y}\) . Assuming that no rubidium atoms have been formed since, what percentage of rubidium atoms were \(^{87} \mathrm{Rb}\) when our solar system was formed \(4.6 \times 10^{9} \mathrm{y}\) ago?
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