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Chapter 13: Problem 52

The nucleus \({ }_{8}^{15} \mathrm{O}\) decays by electron capture. Write (a) the basic nuclear process and (b) the decay process. (c) Determine the energy of the neutrino. Disregard the daughter's recoil.

Short Answer

Expert verified
The basic process for electron capture is \( p + e^- \rightarrow n + \nu \). The decay process for Oxygen-15 is \( ^{15}_{8}O + e^- \rightarrow ^{15}_{7}N + \nu \). The energy of the neutrino is obtained by calculating \( E = (m_{O} - m_{N})c^2 \), where \( m_{O} \) and \( m_{N} \) are the atomic masses of Oxygen-15 and Nitrogen-15, respectively, and \( c \) is the speed of light.

Step by step solution

01

Writing the basic nuclear process

In electron capture, a proton-rich nucleus absorbs an inner shell electron, converts a proton into a neutron, and simultaneously ejects a neutrino. This can be represented as follows: \[ p + e^- \rightarrow n + \nu \] Here, \( p \) denotes a proton, \( e^- \) is an electron, \( n \) is a neutron and \( \nu \) is a neutrino.
02

Writing the decay process

Now, the exercise specifies the nucleus of Oxygen-15, so we indicate that \[ ^{15}_{8}O + e^- \rightarrow ^{15}_{7}N + \nu \] Here, we note that the atomic number (the subscript) decreases by 1, while the atomic mass number (the superscript) remains unchanged. This corresponds to the conversion of a proton into a neutron.
03

Calculating the energy of the emitted neutrino

The energy of the neutrino can be determined from the difference in atomic mass between the initial oxygen atom and the final nitrogen atom. The energy released is completely carried away by the neutrino, since the recoil of the daughter nucleus is disregarded. So, the energy is given by: \[ E = (m_{O} - m_{N})c^2 \] Where \( m_{O} \) is the mass of Oxygen-15, \( m_{N} \) is the mass of Nitrogen-15, and \( c \) is the speed of light. Look these values up in a data table and carry out the calculation to find the energy.

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Key Concepts

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

Electron Capture
In nuclear physics, electron capture is a fascinating process where an unstable nucleus absorbs an orbiting electron. This primarily happens with inner-shell electrons, which are close to the nucleus. When such an electron is captured, it combines with a proton in the nucleus, leading to a fascinating outcome: the proton transforms into a neutron.
  • This process results in a decrease in the atomic number by one, as one proton is effectively removed from the nucleus.
  • The mass number remains unchanged because a neutron, having roughly the same mass as a proton, replaces it.
  • The capture typically results in the emission of a neutrino, a tiny, nearly massless particle, which is shot out from the nucleus.
Understanding electron capture can help explain changes within atoms, especially those moving towards greater stability.
Neutrino Energy Calculation
Following the electron capture in our example with i.e., when is less than parachute weight at collisional calculations , the calculation of neutrino energy becomes crucial. In this process, a significant clue comes from the conservation of energy. The difference in mass between the parent and daughter nucleus represents the energy available in the decay.
  • This energy difference is mostly carried off by the neutrino, as the recoil of the daughter nucleus is often negligible.
  • The formula to calculate this energy is: , where , is the mass of the original nucleus, is the mass of the resulting nucleus after decay, and denotes the speed of light in a vacuum.
Calculating this energy is important because it tells us how much energy is being carried away by the neutrino in the form of kinetic energy. Bringing more clarity to energy conservation helps solidify the understanding of the decay process.
Nuclear Decay Process
A nuclear decay process is a fundamental part of nuclear physics, allowing unstable nuclei to achieve stability. When it comes to electron capture, the nuclear decay process involves the absorption of an electron by the nucleus, specifically a proton.
  • In our oxygen-to-nitrogen example, the nucleus of transitions to by capturing an electron.
  • The atomic number decreases by one because of the proton-to-neutron conversion, completing the transition.
  • A neutrino is emitted as a by-product, being a key evidence of the decay having occurred.
Nuclear decay transforms elements, playing a crucial role in nuclear reactions and the mechanisms of radioactive decay.
Proton to Neutron Conversion
Proton to neutron conversion is a central theme in electron capture. This conversion is a type of beta decay process whereby a proton in an atom's nucleus captures an electron and turns into a neutron.
  • This transformation reduces the atomic number, symbolizing a shift in the identity of the element, such as the conversion from oxygen-15 to nitrogen-15.
  • Crucially, the neutron now makes the nucleus more stable, which is often the driving force behind the transformation.
  • The process maintains the mass number due to the relatively equal mass of protons and neutrons.
Proton to neutron conversion is a simple yet powerful concept once grasped, offering a window into how atomic structures can change to favor stability, and illustrating the dynamic nature of nuclear reactions.

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Most popular questions from this chapter

Using a reasonable scale, sketch an energy-level diagram for (a) a proton and (b) a deuteron, both in a magnetic field \(B\). (c) What are the absolute values of the changes in energy that accompany the possible transitions between the levels shown in your diagrams?

During the manufacture of a steel engine component, radioactive iron \(\left({ }^{59} \mathrm{Fe}\right)\) is included in the total mass of \(0.2 \mathrm{~kg}\). The component is placed in a test engine when the activity due to this isotope is \(20 \mu\) Ci. After a 1000 -h test period, oil is removed from the engine and found to contain enough \({ }^{59} \mathrm{Fe}\) to produce 800 disintegrations \(/\) min per liter of oil. The total volume of oil in the engine is \(6.5 \mathrm{~L} .\) Calculate the total mass worn from the engine component per hour of operation. (The halflife for \({ }^{59} \mathrm{Fe}\) is \(45.1\) days.)

Find the radii of (a) a nucleus of \({ }_{2}^{4} \mathrm{H}\) and (b) a nucleus of \({ }_{92}^{238} \mathrm{U} .\) (c) What is the ratio of these radii?

As part of his discovery of the neutron in 1932 , James Chadwick determined the mass of the newly identified particle by firing a beam of fast neutrons, all having the same speed, at two different targets and measuring the maximum recoil speeds of the target nuclei. The maximum speeds arise when an elastic head-on collision occurs between a neutron and a stationary target nucleus. (a) Represent the masses and final speeds of the two target nuclei as \(m_{1}, v_{1}, m_{2}\), and \(v_{2}\) and assume Newtonian mechanics applies. Show that the neutron mass can be calculated from the equation $$ m_{n}=\frac{m_{1} v_{1}-m_{2} v_{2}}{v_{2}-v_{1}} $$ (b) Chadwick directed a beam of neutrons (produced from a nuclear reaction) on paraffin, which contains hydrogen. The maximum speed of the protons ejected was found to be \(3.3 \times 10^{7} \mathrm{~m} / \mathrm{s}\). Since the velocity of the neutrons could not be determined directly, a second experiment was performed using neutrons from the same source and nitrogen nuclei as the target. The maximum recoil speed of the nitrogen nuclei was found to be \(4.7 \times 10^{6} \mathrm{~m} / \mathrm{s}\). The masses of a proton and a nitrogen nucleus were taken as \(1 \mathrm{u}\) and \(14 \mathrm{u}\), respectively. What was Chadwick's value for the neutron mass?

(a) Find the radius of the 12 C nucleus. (b) Find the force of repulsion between a proton at the surface of a \({ }^{12}{6}\) C nucleus and the remaining five protons. (c) How much work (in MeV) must be done to overcome this electrostatic repulsion and put the last proton into the nucleus? \((\mathrm{d})\) Repeat \((\mathrm{a}),(\mathrm{b})\), and (c) for \({ }_{92}^{238} \mathrm{U}\).
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