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A series of Mössbauer experiments is performed with the same emitter and absorber but with the emitter placed in various host materials. The absorber is always in the same host. (a) Show that the chemical shift (the absorber velocity corresponding to the center of the spectrum) is a linear function of the electron probability density \(\rho\) at the site of the emitter and so is given by \(v=a \rho+b\), where \(a\) and \(b\) do not depend on the sample in which the emitter is placed. (b) The following data was recorded for four samples : \(v_{1}=\) \(1.42 \mathrm{~mm} / \mathrm{sec}, v_{2}=0.23 \mathrm{~mm} / \mathrm{sec}, v_{3}=0.37 \mathrm{~mm} / \mathrm{sec}\), and \(v_{4}=0.95 \mathrm{~mm} / \mathrm{sec} .\) For the first two samples \(\rho\) was found using other experimental data, with the results \(\rho_{1}=8.0248 \times\) \(10^{34} \mathrm{~m}^{-3}\) and \(\rho_{2}=8.0286 \times 10^{34} \mathrm{~m}^{-3}\), respectively. Find the values of \(a\) and \(b\), then find the electron probability densities for samples 3 and 4 .

Short Answer

Expert verified
The coefficients 'a' and 'b' for the linear function can be calculated using the given two sets of data regarding observer velocity and electron probability density. After the derivation of these two coefficients, they can be further used to find the electron probability densities for samples 3 and 4 using the corresponding velocities.

Step by step solution

01

Understand and Validate the given data

From the question we have four sets of velocity data (emitter absorber velocities): \(v_{1}=1.42 \mathrm{~mm} / \mathrm{sec}\), \(v_{2}=0.23 \mathrm{~mm} / \mathrm{sec}\), \(v_{3}=0.37 \mathrm{~mm} / \mathrm{sec}\), \(v_{4}=0.95 \mathrm{~mm} / \mathrm{sec}\). Here, we also have two sets of electron probability densities: \(\rho_{1}=8.0248 \times 10^{34} \mathrm{~m}^{-3}\) , \(\rho_{2}=8.0286 \times 10^{34} \mathrm{~m}^{-3}\). The task is to find \(a\) and \(b\) which originate from the linear function of chemical shift \(v=a \rho+b\).
02

Calculate 'a' and 'b' from given data

Given two points \((\rho_1, v_1)\) and \((\rho_2, v_2)\), we can calculate the slope of the line (a) using the formula \(a = \frac{\Delta v}{\Delta \rho} = \frac{v_2 - v_1}{\rho_2 - \rho_1}\). Once we have calculated 'a', we can put this value and one of the data points into our linear function to solve for 'b': \(b = v - a * \rho\).
03

Find electron probability densities for samples 3 and 4

Now that we have 'a' and 'b', we can use them to find the electron probability densities \(\rho_3\) and \(\rho_4\) using their corresponding velocities \(v_{3}=0.37 \mathrm{~mm} / \mathrm{sec}\) and \(v_{4}=0.95 \mathrm{~mm} / \mathrm{sec}\), respectively. Using the formula \(\rho = \frac{v - b} {a}\), we can solve for \(\rho_{3}\) and \(\rho_{4}\).

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Key Concepts

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

Chemical Shift
When studying the concept of chemical shift via Mössbauer spectroscopy, one dives into the intricate details of how the atomic nucleus interacts with its electron environment. Chemical shift occurs due to the electrons surrounding the nucleus and their influence on the nuclear energy levels, effectively shifting the resonance energy required for gamma-ray absorption.

Within this spectroscopic technique, the chemical shift is sensitive to changes in the electron density around the nucleus. In the Mössbauer experiment outlined in the exercise, the chemical shift is measured by the absorber velocity where the gamma-ray absorption peaks. It is crucial to understand that the chemical shift acts as a probe, offering insights into the electronic environment and chemical structure of the absorber material.

Impact on Chemical Shift

The greater the electron probability density \(\rho\) near the nucleus, the larger the shift. This is because the electrons exert a shielding effect on the nucleus from the external electromagnetic fields. This effect is quantifiable and, as the exercise suggests, can be expressed as a linear relationship \( v = a\rho + b \). In a series of experiments with varied emitter materials but a constant host for the absorber, the dependency of chemical shift on \(\rho\) proves fundamental in interpreting Mössbauer spectra.
Electron Probability Density
Digging deeper into the realm of physics, the electron probability density \(\rho\) is, in essence, a mathematical expression that describes the likelihood of finding an electron in a particular region around the nucleus. In Mössbauer spectroscopy, this concept is paramount as it dictates the degree of interaction between the nucleus and the electrons, which in turn affects the chemical shift we observe.

In the provided exercise, electron probability density is not an abstract concept; it is a measurable entity that can be determined through experimental techniques other than Mössbauer spectroscopy. By correlating the absorber velocity (chemical shift) to the known electron probability density of certain samples, we can deduce a linear relationship that allows for the calculation of unknown \(\rho\) values.

Practical Application

In practical terms, knowing \(\rho\) for different samples enables us to understand and predict how atomic nuclei in those materials will respond to gamma rays in a Mössbauer experiment. The exercise illustrates how, by using only two known \(\rho\) values and their corresponding absorber velocities, we can deduce the electron probability densities for additional samples with unknown \(\rho\) values.
Linear Function
Mathematics often provides elegant solutions to physical phenomena, and in the context of Mössbauer spectroscopy, a linear function is used to describe the relationship between chemical shift and electron probability density. A linear function is a straight-line equation of the form \( y = mx + c \) where \( m \) is the slope and \( c \) is the y-intercept.

In our exercise, \( v = a\rho + b \) is a linear equation where \( v \) represents the chemical shift, \( a \) the slope—which corresponds to how much \( v \) changes with \( \rho \)—and \( b \) the y-intercept. By determining the slope \( a \) and the intercept \( b \) from the given data, we create a predictive model that can be used to calculate the unknown electron probability densities for other samples.

Understanding Linearity

The idea that this relation is linear means that as the electron probability density increases, the chemical shift does so at a constant rate, which is a pivotal point in understanding the behavior of nuclear resonance in different environments. Thus, the exercise emphasizes the simplicity and power of using basic algebraic functions to grasp more complex physical interactions.

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Most popular questions from this chapter

Explain exactly why the optical model potential which a nucleus exerts on a bombarding nucleon of energy \(50 \mathrm{MeV}\) is different from the shell model potential which it exerts on one of its own nucleons. What would you expect the optical model potential to be like for a bombarding nucleon of energy \(5 \mathrm{MeV}\) ?

In considering the effects of radiation on the human body, it is necessary to define units for the amount of radiation absorbed. One of these is the rad (radiation absorbed dose): 1 rad indicates an average of \(0.01\) joule of absorbed energy per \(\mathrm{kg}\) of body tissue, regardless of which part of the body actually was exposed. A \(75 \mathrm{~kg}\) worker at a hospital radiology lab inadvertently swallows a capsule containing \(5 \mathrm{mg}\) of \({ }^{88} \mathrm{Ra}^{226}\) (half-life \(=1600\) years). This isotope of radium undergoes alpha-decay, each \(\alpha\) particle carrying an energy of \(4.87 \mathrm{MeV}\). If \(90 \%\) of these particles are stopped inside the man's body, what radiation dose does he receive in 12 hours?

(a) Use (15-16) with \(Q=0\) to calculate the energy lost by a \(1 \mathrm{MeV}\) fission neutron to the recoil of \({ }^{6} \mathrm{C}^{12}\), if it scatters elastically at the typical angle \(90^{\circ}\) from such a nucleus in the moderator of a nuclear reactor. (b) How much energy does it lose in a \(90^{\circ}\) scattering if its energy has been reduced to \(0.001 \mathrm{MeV} ?\) (c) How much energy does it have, on the average, if it is in thermal equilibrium at an operating temperature of \(500^{\circ} \mathrm{K}\) ? (d) Estimate the number of scatterings required to bring the neutron into thermal equilibrium.

A hypothetical \(\mathrm{H}\)-bomb with the explosive power of 50 Megatons of TNT uses the reaction $${ }^{1} \mathrm{H}^{2}+{ }^{1} \mathrm{H}^{2} \rightarrow{ }^{2} \mathrm{He}^{3}+{ }^{0} n^{1}$$ (Atomic masses are: \(\mathrm{H}^{2}, 2.014102 u ; \mathrm{He}^{3}, 3.016029 u\).) The required A-bomb "trigger" is rated at 2 Megatons (included in the 50 above). One ton of TNT produces \(2.6 \times 10^{22}\) \(\mathrm{MeV}\) of energy. (a) How much energy does each fusion produce? (b) How much hydrogen does the bomb contain?

\({ }^{26} \mathrm{Fe}^{57}\), in a ferromagnetic iron sample, is used as an emitter in a Mössbauer experiment. The absorber is in stainless steel and has a single narrow Mössbauer peak in its absorption spectrum. The emitter is in a steady magnetic field so the first excited state splits into 4 levels, identified by \(m_{\mathrm{e}}=-3 / 2,-1 / 2,+1 / 2\), or \(+3 / 2\), while the ground state splits into 2 levels, identified by \(m_{\mathrm{g}}=-1 / 2\) or \(+1 / 2 .\) The energies of the excited states are given by \(E_{\mathrm{c}}+2 \mu_{\mathrm{c}} \mathrm{Bm}_{\mathrm{e}} / 3\) and those for the ground states are given by \(E_{\mathrm{g}}-2 \mu_{\mathrm{g}} \mathrm{Bm}_{\mathrm{g}}\), where \(E_{\mathrm{e}}\) and \(E_{\mathrm{g}}\) are the energies in the absence of a magnetic field. The magnetic dipole moments of the states are \(\mu_{\mathrm{e}}\) and \(\mu_{\mathrm{g}}\), respectively. The signs in the energy equations are different because the moments are in opposite directions for the excited and ground states. (a) Neglect any chemical shift and show that the Mössbauer peaks occur for absorber velocities given by $$ v\left(m_{e} \rightarrow m_{\mathrm{g}}\right)=-\frac{c B}{E_{e}-E_{\mathrm{g}}}\left(\frac{2}{3} \mu_{\mathrm{e}} m_{\mathrm{e}}+2 \mu_{\mathrm{g}} m_{\mathrm{g}}\right) $$ (b) Show that the ratio of the magnetic dipole moments is $$ \frac{\mu_{\mathrm{e}}}{\mu_{\mathrm{g}}}=3 \frac{v(3 / 2 \rightarrow 1 / 2)-v(1 / 2 \rightarrow 1 / 2)}{v(1 / 2 \rightarrow 1 / 2)-v(1 / 2 \rightarrow-1 / 2)} $$ (c) Once the chemical shift is subtracted, typical experimental values are \(v(3 / 2 \rightarrow 1 / 2)=\) \(-5.57 \mathrm{~mm} / \mathrm{sec}, v(1 / 2 \rightarrow 1 / 2)=-3.14 \mathrm{~mm} / \mathrm{sec}\), and \(v(1 / 2 \rightarrow-1 / 2)=+1.04 \mathrm{~mm} / \mathrm{sec}\). Calculate the magnetic dipole moment ratio and the magnetic field at the site of the emitter. Take \(\mu_{\mathrm{g}}=4.56 \times 10^{-28}\) joule/tesla.

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