/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 Some radionuclides decay by capt... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Some radionuclides decay by capturing one of their own atomic electrons, a \(K\) -shell electron, say. An example is $${ }^{49} \mathrm{~V}+\mathrm{e}^{-} \rightarrow{ }^{49} \mathrm{Ti}+\nu, \quad T_{1 / 2}=331 \mathrm{~d}$$ Show that the disintegration energy \(Q\) for this process is given by $$Q=\left(m_{\mathrm{V}}-m_{\mathrm{Ti}}\right) c^{2}-E_{K}$$ where \(m_{\mathrm{v}}\) and \(m_{\mathrm{Ti}}\) are the atomic masses of \({ }^{49} \mathrm{~V}\) and \({ }^{49} \mathrm{Ti},\) respectively, and \(E_{K}\) is the binding energy of the vanadium \(K\) -shell electron. (Hint: Put \(\mathbf{m}_{\mathrm{V}}\) and \(\mathbf{m}_{\mathrm{Ti}}\) as the corresponding nuclear masses and then add in enough electrons to use the atomic masses.)

Short Answer

Expert verified
The disintegration energy is \(Q = (m_V - m_{Ti})c^2 - E_K\).

Step by step solution

01

Understanding the Reaction

The process involves capturing an atomic electron by \({ }^{49} \text{V}\) leading to the formation of \({ }^{49} \text{Ti}\) and releasing a neutrino \(u\). This electron capture changes the number of protons, transforming vanadium into titanium.
02

Define Masses Involved

The reaction can be written as \( { }^{49} \text{V} + e^− \to { }^{49} \text{Ti} + u \). Here, \(m_V\) is the atomic mass of vanadium, and \(m_{Ti}\) is the atomic mass of titanium. We need to express the disintegration energy \( Q \) in terms of these atomic masses.
03

Relate Nuclear to Atomic Masses

The nuclear masses are \( m_V' \) and \( m_{Ti}' \), the real atomic masses are related to these by \( m_{V} = m_V' + Z_Vm_e - E_K/c^2 \) and \( m_{Ti} = m_{Ti}' + Z_{Ti}m_e \). Here \( m_e \) is the electron mass, \( E_K \) is the binding energy of the K-shell electron, and \( Z_{V}, Z_{Ti} \) represent the atomic numbers.
04

Calculating Disintegration Energy \(Q\)

The energy released \(Q\) is the mass difference, \( Q = \left(m_V' + Z_Vm_e - E_K/c^2 \right) - \left(m_{Ti}' + Z_{Ti}m_e\right) \) times \(c^2\). Simplicity shows that \( Z_V = Z_{Ti} + 1 \), leading to \( Q = \left(m_V - m_{Ti}\right) c^2 - E_K.\) This aligns with how the atomic electron mass and nuclear potential reduce.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Binding Energy
Binding energy is a key concept when discussing atomic structure and nuclear reactions, like electron capture. In an atom, electrons are arranged in shells around the nucleus, with the innermost shell being the K-shell. Each electron is bound to the nucleus with a particular energy known as the binding energy.
The binding energy plays a crucial role in nuclear decay processes. In electron capture, a K-shell electron is captured by the nucleus. This particular process absorbs the binding energy of the electron.
  • This energy must be considered when calculating the disintegration energy.
  • In our example, the binding energy is subtracted from the total energy release, as the K-shell electron is assumed to be absorbed.
Understanding binding energy helps in predicting the stability of a nucleus and the likelihood of specific decay modes, like electron capture in certain radionuclides. It contributes to the overall energy balance in nuclear reactions.
Disintegration Energy
Disintegration energy, often denoted by the symbol \(Q\), refers to the energy released during a nuclear decay process such as electron capture. It is the difference in mass-energy between the original and resulting nuclei, adjusted by any associated electron binding energies.
  • The disintegration energy is a crucial factor in determining the behavior of nuclear decays, as it reflects the changed binding energy from before and after the reaction.
  • In the exercise, the equation \(Q = \left(m_{\text{V}} - m_{\text{Ti}}\right)c^2 - E_{K}\) illustrates how to account for the captured electron's binding energy when calculating the energy released.
To find the disintegration energy, one must consider the transformed atomic masses including any changes through electron capture. The mass difference is multiplied by the speed of light squared \(c^2\) to convert into an energy equivalent (from mass to energy - Einstein's famous \(E=mc^2\)). This calculated energy tells us how much energy will be liberated or absorbed during the decay.
Nuclear Decay
Nuclear decay, also known as radioactive decay, is a natural process where an unstable atomic nucleus loses energy by emitting radiation. The important concepts linked to nuclear decay include:
  • Transmutation of elements: In electron capture, as detailed in this example, a K-shell electron is captured by the nucleus of an atom, transforming it into a different element.
  • Stability factors: This process often occurs when the original atom is more unstable compared to the resulting atom. The electron capture process stabilizes the atom by converting a proton into a neutron, usually when additional neutrons can lead to a more stable nucleus.
  • Released particles: In electron capture, a neutrino (often denoted \(u\)) is released. This process is one form of nuclear decay, differing from others like alpha or beta decay where particles are expelled from the atom.
Nuclear decay like electron capture is a key way in which elements change form over time, releasing energy and forming new elements. Understanding it helps scientists in fields ranging from nuclear physics to astrophysics as they study the lifespan of stars and the formation of elements.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When aboveground nuclear tests were conducted, the explosions shot radioactive dust into the upper atmosphere. Global air circulations then spread the dust worldwide before it settled out on ground and water. One such test was conducted in October 1976 . What fraction of the \({ }^{90} \mathrm{Sr}\) produced by that explosion still existed in October 2006 ? The half-life of \({ }^{90} \mathrm{Sr}\) is \(29 \mathrm{y}\).

The radioactive nuclide \({ }^{99}\) Tc can be injected into a patient's bloodstream in order to monitor the blood flow, measure the blood volume, or find a tumor, among other goals. The nuclide is produced in a hospital by a "cow" containing \({ }^{99} \mathrm{Mo},\) a radioactive nuclide that decays to \({ }^{99} \mathrm{Tc}\) with a half-life of \(67 \mathrm{~h}\). Once a day, the cow is "milked" for its \({ }^{99} \mathrm{Tc},\) which is produced in an excited state by the \({ }^{99} \mathrm{Mo} ;\) the \({ }^{99} \mathrm{Tc}\) de- excites to its lowest energy state by emitting a gamma-ray photon, which is recorded by detectors placed around the patient. The de-excitation has a half- life of \(6.0 \mathrm{~h}\). (a) By what process does \({ }^{99}\) Mo decay to \({ }^{99} \mathrm{Tc} ?\) (b) If a patient is injected with an \(8.2 \times 10^{7}\) Bq sample of \({ }^{99} \mathrm{Tc}\), how many gamma-ray photons are initially produced within the patient each second? (c) If the emission rate of gamma-ray photons from a small tumor that has collected \({ }^{99} \mathrm{Tc}\) is 38 per second at a certain time, how many excited-state \({ }^{99} \mathrm{Tc}\) are located in the tumor at that time?

A typical chest \(x\) -ray radiation dose is \(250 \mu \mathrm{Sv},\) delivered by x rays with an RBE factor of \(0.85 .\) Assuming that the mass of the exposed tissue is one-half the patient's mass of \(88 \mathrm{~kg}\), calculate the energy absorbed in joules.

The half-life of a particular radioactive isotope is \(6.5 \mathrm{~h}\). If there are initially \(48 \times 10^{19}\) atoms of this isotope, how many remain at the end of \(26 \mathrm{~h}\) ?

An \(85 \mathrm{~kg}\) worker at a breeder reactor plant accidentally ingests \(2.5 \mathrm{mg}\) of \({ }^{239} \mathrm{Pu}\) dust. This isotope has a half- life of \(24100 \mathrm{y}\), decaying by alpha decay. The energy of the emitted alpha particles is \(5.2 \mathrm{MeV},\) with an \(\mathrm{RBE}\) factor of \(13 .\) Assume that the plutonium resides in the worker's body for \(12 \mathrm{~h}\) (it is eliminated naturally by the digestive system rather than being absorbed by any of the internal organs) and that \(95 \%\) of the emitted alpha particles are stopped within the body. Calculate (a) the number of plutonium atoms ingested, (b) the number that decay during the \(12 \mathrm{~h},\) (c) the energy absorbed by the body, (d) the resulting physical dose in grays, and (e) the dose equivalent in sieverts.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.