The Mössbauer effect is a tool which can be used to study the environments in which nuclear absorbers and emitters are imbedded. In this experiment, we use radioactive ${}^{57}$Co to produce ${}^{57}$Fe in an excited nuclear state. The 14.4 keV gamma emitted by ${}^{57}$Fe as it relaxes to the ground state can then be used to excite ${}^{57}$Fe atoms in various absorber materials, and the absorption spectra will provide information about the material.

In particular, the objectives of this experiment are as follows:

- to understand the physical origin of the Mössbauer effect including the following:
- what is a phonon and how does the phonon density of states lead to recoilless emission and absorption?
- how can the Debye model be used to predict the recoilless fraction? and
- why is the 14.4 keV line of ${}^{57}$Fe used for Mössbauer spectroscopy?

- to understand the apparatus necessary for Mössbauer spectroscopy, including the following:
- how does a moving radioactive source lead to a “tunable” photon source via Doppler shift?
- how can we calibrate our source via interferometry? and
- how do we interpret absorption spectra in light of the characteristics of the absorber and emitter?

- to observe and interpret simple recoilless emission and absorption of the 14.4 keV transition photon of ${}^{57}$Fe; and
- to use Mössbauer spectroscopy to study perturbational effects in absorbers, including the following:
- to observe the isomer shift due to differing emitter and absorber environments;
- to observe the splitting of the single transition line into six transitions due to the Zeeman effect in an enriched ${}^{57}$Fe sample, and to measure the magnitude of the energy level splitting and to therefore determine the local magnetic field and the magnetic moment of the $I = 3/2$ excited state;
- to observe the splitting of the single transition line into two transitions due to quadrupole splitting in a nitroprusside sample, and to measure the magnitude of the energy level splitting in order to estimate the local electric field gradient;
- to observe the temperature-dependent shift due to the second-order Doppler effect, and to compare the measurements of this effect to predictions in order to estimate the Debye temperature of stainless steel; and
- to observe the absorber spectrum of MetGlas at different temperatures, and to quantitatively compare it to other observed spectra.

Mössbauer spectroscopy can serve as a way of non-destructively gathering information about the state of iron present in a sample. For instance:

- The mars rovers Spirit and Opportunity ( NASA Link) have Mossbauer spectrometers that are used to probe the martian soil for gather informations about its iron contents.
- In 2004 Spirit found evidence of the mineral Goethite on mars, which was evidence for the presence of water in the past.
- Mossbauer spectroscopy has been used to Identify counterfit bills by detecting unusual amounts of iron compounds in the ink.
- Archeologists have used this technique to gather information about pottery glazes. The iron compounds formed in the glazes are dependent on factors such as the oxygen content of the kiln and the temperature.
- A 2018 paper in Nature uses Mossbauer spectroscopy to probe a thin film of single molecule magnets (SMMs), which are a candidate for the basis of a quantum computing system.

- [2] H. Frauenfelder,
*The Mossbauer Effect*, W. A. Benjamin, Inc., 1962. - [8] N. N. Greenwood and T. C. Gibb,
*Mossbauer Spectroscopy*, Chapman and Hall Ltd., London, 1971. - [9] E. U. Condon and G. H. Shortley,
*The Theory of Atomic Spectra*, Cambridge University Press, New York, 1935. - [10] H. Kopferman
*Nuclear Moments*, Academic Press Inc., 1958. - [12] Y.-L. Chen and D.-P. Yang,
*Mossbauer Effect in Lattice Dynamics*, Wiley VCH, 2007.

Read over the theory section below and and answer the following in preparation for your first day in lab:

(a) The true transition energy of the excited state of ${}^{57}$Fe we will be examining is 14.41300 ± 0.00015 keV. What is the magnitude of the photonenergy shift due to recoilof the ${}^{57}$Fe? (Don't worry about uncertainty.)

(b) The nominal value for the lifetime of the transition we will be examining is approximately 140 ns. What is thenatural linewidthof this state? How does the linewidth compare to the photon energy shift due to recoil? How does the linewidth compare to the energy of the transition?

(c) In principle, a hydrogen atom emitting a photon due to electronic transitions will recoil as well. The Lyman-α line of the hydrogen spectrum is caused by the decay of a hydrogen atom from an $n = 2$ state to the ground state ($n = 1$), with an energy of 10.2 eV and a mean lifetime of 1.6 ns. Will recoil effects be significant compared to the natural linewidth of this transition?

(d) What is the expected energy Doppler shift for a source emitting 14.4 keV gamma rays in the source’s frame and moving with a velocity of 1 cm/s relative to the lab frame?

An atomic nucleus consists of a large number of nucleons (protons and neutrons) interacting primarily by the strong nuclear force. As with other bound quantum mechanical systems – such as the harmonic oscillator or the hydrogen atom – the laws of quantum mechanics lead to this system having a ground state and a set of discrete excited states at higher energies. Moreover, as in the case of a hydrogen atom, the nucleus can be excited to a higher state via absorption of photons of the appropriate energy, and can decay from an excited state to a lower state by emitting a photon. Typical energies of nuclear transitions range from tens of keV to tens of MeV, compared to tens of electron-volts for typical atomic transitions.

In principle, a $\gamma$-ray photon emitted from a nucleus decaying from some excited state to the ground state could then travel to a second, identical nucleus and boost it up to the original excited state of the first nucleus; these energies should, in theory, be equal. However, in practice this symmetry is destroyed by the effects of *nuclear recoil.* (See Fig.1.) A photon of energy $E_\gamma$ emitted by a decaying nucleus will necessarily have a momentum given by $p_\gamma = E_\gamma / c$; in order to conserve momentum, then, the nucleus will have to recoil with an equal and opposite momentum. This means that some of the energy of the decay will go into the kinetic energy of the nucleus, and as a consequence the energy of the emitted photon will be less than the energy of the transition. Similarly, a photon which is absorbed by a nucleus (initially at rest) must have enough energy to both excite the nucleus *and* impart its momentum to the nucleus.

The net result is that a photon emitted by a decaying nucleus, initially at rest, will have *less* energy than the true transition energy, while a photon which can be absorbed by a ground-state nucleus at rest must have *more* energy than the true transition energy. Quantitatively, we can work out this energy shift by considering conservation of energy and momentum. If a nucleus at rest decays, emitting a photon, conservation of momentum requires that

$\dfrac{E_{\gamma}}{c}-p_n = 0$, | (1) |

where $p_n$ is the momentum of the nucleus after the collision. Assuming that the energy of the transition is much less than the rest-energy of the nucleus (i.e.$E_\gamma << mc^2$, with *m* the mass of the nucleus), the kinetic energy of the nucleus after the decay will be given by $p_n^2 /2m$. Conservation of energy then requires that

$E_t = E_\gamma + \dfrac{p^2_n}{2m}$, | (2a) |

where $E_t$ is the “true transition energy”, i.e., the energy difference between the excited and ground states. Solving these two equations, we can show that the energy of the photon will be approximately given by

$E_\gamma \approx E_t -\dfrac{E^2_t}{2mc^2}$. | (2b) |

A similar calculation holds for the absorption of a photon by a recoiling nucleus. The result is that the energy shift of the photon is of the same magnitude as in the case of emission, but of opposite sign (positive instead of negative.)

This shifting of the emission and absorption photon energies can be sufficiently extreme that a typical photon emitted by an excited nucleus at rest decaying to the ground state will not be absorbed by another nucleus at rest in the ground state. Each nuclear transition has a characteristic linewidth $\Gamma$, related to its mean lifetime $\tau$ by

$\Delta E \Delta t = (\Gamma /2)\tau = \hbar /2$. | (3) |

If the shift of the photon energies due to recoil is large compared to the natural linewidth of the transition, a typical photon from a decay will not be able to excite another ground-state nucleus at rest. (See Fig. 2a.)

If absorption is to occur, then both the emission and absorption must occur without recoil (i.e., the emission must be *recoilless*). If the emitting (or absorbing) nucleus is embedded in the crystal lattice of a solid, there can be a significant probability that the entire lattice recoils, rather than a single nucleus. In this case, the factor of $m$ in Eq. (3) will effectively be replaced by $Nm$, where $N$ is the number of nuclei that recoil. Since this number can be very large, the photon energy shift due to recoil will be many orders of magnitude smaller than in the case of a free atom, and the absorption and emission spectra will overlap. (See Fig. 2b.) This effect, in which atoms in a solid sample can absorb gamma rays without an energy shift due to recoil, is named the *Mössbauer effect* after its discoverer Rudolf Mössbauer.

We have seen how recoilless reactions allow us to excite nuclei which are at rest and otherwise unperturbed. By itself, this technique would be interesting but not terribly useful; what would be even more useful would be a “tunable” source of gamma rays near this transition, so that we could observe how these energy levels are perturbed when the nucleus is exposed to different environments. We can accomplish this tuning by moving the emitting nucleus relative to the absorbing nucleus and thereby Doppler-shifting the gamma rays that the absorbers “see”. (See Fig. 2c.) Quantitatively, if our source emits a photon of energy $E_t$ in the source’s frame, and the source is moving with velocity $v \ll c$ with respect to the lab frame, then the energy of the photon in the lab frame will be

$E_t' = E_t + \dfrac{v}{c}E_t$. | (4) |

Thus, by moving our source at different velocities we can tune the energies of the gamma rays incident on our absorber, as desired.

In the absence of any external effects (like the presence of electric or magnetic fields), the energy difference $\Delta E$ between the $I = 1/2$ and $I = 3/2$ nuclear states of ${}^{57}$Fe is 14.4 keV. (See Fig. 3.)

However, the local environment in which the nucleus is sitting can cause this energy difference to change. For example, consider the following:

- The fields produced by the orbital electrons of the host atom and neighboring atoms can cause energy levels in the nucleus to change, – and may even cause different changes to different energy levels, – leading to isomer shifts.
- If the ${}^{57}$Fe nucleus sits in a magnetic field, both the $I = 3/2$ and $I = 1/2$ states will experience a splitting due to the Zeeman effect, resulting in a change from one transition energy to multiple transition energies.
- Electric field gradients due to the molecular structure surrounding the ${}^{57}$Fe nucleus can split the $I = 3/2$ energy state, leading to the quadrupole effect.

It is these sort of changes to the energy level structure of the nucleus which Mössbauer spectroscopy can study.

The nucleus is not a point of charge, but instead is approximately a sphere of some finite radius, $R$. This radius depends on the energy state that the nucleus is in (e.g., $R_0$ in the ground state need not be the same as $R_1$ in the first excited state), and this radius may even be effected by the local environment in which the nucleus is embedded (e.g., $R_0$ for ${}^{57}$Fe in naturally occurring iron ore need not be the same as $R_0$ for ${}^{57}$Fe in stainless steel). Changes in the transition energy between states due to these effects are *isomer shifts*.

If the absorber and the emitter are embedded in the same chemical environment, then the emission transmission energy will match the absorption transmission energy and no *relative* shift between absorber and emitter will be seen. However, if the absorber and the emitter are embedded in different chemical environments, then the emitted and absorbed energies may differ such that

$\Delta E_a - \Delta E_e = \delta$, | (8) |

where $\delta$ is termed the chemical isomer shift (i.e., the isomer shift due to chemical environment). Figure 4 illustrates a common isomer shift situation. On the left we illustrate a transition of energy $\Delta E$ between two levels of a nucleus, while on the right we illustrate a transition of energy $\Delta E_{iso}$ between two levels of nucleus in a different chemical environment. We see that not only do the absolute energies of the ground and excited states change, but the difference between those states changes by an amount $\Delta E - \Delta E_{iso} = \delta$. The isomer shift can be either positive or negative.

Most nuclei have a magnetic dipole moment $\mu$ due to the magnetic moments of their constituent nucleons and the angular momenta of their protons. In the presence of an external magnetic field $\bf H$, this causes the Hamiltonian to be perturbed as

E $\mathcal{H}_{mag} = -\mu \cdot \mathbf{H} = -\mu_Z \left | \mathbf{H} \right |$, | (5) |

where we have defined the $z$-direction to point along the magnetic field. This perturbation can be expressed in terms of the angular momentum of the nucleus $I$ by the introduction of the $g$-factor for the state,

$\mathcal{H}_{mag} = \dfrac{g \mu_N \left | \mathbf{H} \right |}{\hbar} I_Z$, | (6) |

where $\mu_N = e\hbar/2m_p$ is the nuclear magneton. The states of definite $I_Z$ will then be energy eigenstates of quantum number (eigenvalue) $m_l$ such that $I_Z = m_I\hbar$, and the energy shifts (away from the unperturbed states) will be given by

$\Delta E_{mag} = g \mu_N\left |\mathbf{H}\right | m_I = \dfrac{\mu}{I} \left | \mathbf{H} \right | m_I$. | (7) |

The factor of 1 divided by $I$ in the second equation comes from the fact that the magnetic moment $\mu$ of the nucleus is defined to be the $z$-component of $\boldsymbol{\mu}$ when $I_Z$ is at its maximum value $I$ ; in other words, $\mu_Z = I_Z\mu /I$.

This derivation will probably look familiar to the reader familiar with the atomic Zeeman effect; the only difference thus far is the substitution of the nuclear spin $\bf I$ for the total electron spin $\bf J$ and the nuclear magneton for the Bohr magneton. However, an important difference is hidden in the $g$-factor. In the case of many atomic transitions, we can obtain an approximate numerical value for $g$, in this case called the *Landé $g$-factor*. In the case of a nucleus, however, the magnetic moment is the superposed magnetic moment of dozens of strongly interacting particles; we cannot make the kinds of approximations required to derive the nuclear $g$-factor. Rather, nuclear magnetic moments must be experimentally determined. These values can be found to be either positive or negative.

This splitting of the nuclear energy levels will cause our absorption lines to split in the presence of a magnetic field. We will be examining a transition from a $I = 1/2$ ground state to an $I = 3/2$ excited state. This means that in the presence of a magnetic field, the ground state will split into two states and the excited state will split into four. Therefore, one might expect there to be eight transition lines in the absorption spectrum – one for each ground state going into each excited state – but the usual selection rules on the change in $m_l$ ($\Delta m_l = +1,0,-1$) will apply, leaving us with only six allowed transitions. (See Fig 5.)

The differences between adjacent split energy states (see Fig. 5) are given by

$\Delta E_1 = \mu_1 \left | \mathbf{H}\right | /I_1$ | (8) |

and

$\Delta E_0 = \mu_0 \left | \mathbf{H}\right | /I_0$, | (9) |

where $\mu_0$ is the magnetic moment of the ground state and $\mu_1$ is the magnetic moment of the excited state. The magnetic moment of the ground state of ${}^{57}$Fe has been measured via electron-nuclear double resonance to be $\mu_0 = (0.09024 \pm 0.00007)\mu_N$[11]; thus, we know that $\Delta E_0$ is positive. However, the shape of the spectrum will still depend on the relative magnitude and signs of $\Delta E_0$ and $\Delta E_1$. By measurements of these energies, we can determine the magnetic moment $\mu_1$ of the excited state and the magnitude of the magnetic field $|\bf{H}\rm |$ in the vicinity of the nucleus.

**NOTE**: By considering relative intensities and comparing different expectations under different relative sign and magnitude combinations, one can infer from the Mössbauer spectrum that $\mu_1$, the moment of the excited state, must be negative and that therefore the energy shift, $\Delta E_1$, is also negative. We will not ask you to work through this logic in this experiment and may instead assume from the start that $\mu_1$ is negative.

We saw above that when a nucleus is placed inside a magnetic field, previously degenerate energy states split into states of different energies. Another method for splitting degenerate levels is to place the nucleus in an environment with an electric field gradient. If the charge distribution of the nucleus is non-symmetric, then there is an interaction between the electric field gradient and the nuclear moment $I$, and energy levels can again be shifted up or down according to their orientation, $m_I$. The magnitude of this interaction is proportional to the quadrupole moment of the energy state, $Q$, which measures the departure from spherical symmetry; a perfect sphere of charge has $Q = 0$, an ellipsoid with the long axis parallel to the quantization axis $z$ has $Q > 0$, and an ellipsoid with the long axis perpendicular to $z$ has $Q < 0$. (See Fig. 7.)

The ground state of ${}^{57}$Fe has nuclear moment $I = 1/2$ and is spherically symmetric; therefore, it is unaffected by any electric field gradient. The excited state has $I = 3/2$ and has an oblong charge distribution; this state will be split. The $I = 3/2$ state has an intrinsic quadrupole moment $\tilde{Q}$, but the relative orientation of this moment with respect to the direction of the field gradient is set by the orientation of the nuclear moment, $m_I$. Without derivation, the effective nuclear quadrupole moment (measured with respect to the electric field gradient as the quantization axis) is given by

$Q = \dfrac{3m_I^2 - I(I+1)}{I(2I-1)}\tilde{Q}$ | (21) |

and the total energy shift (away from the unperturbed state) is given by

$\Delta E = \dfrac{1}{4}eQ\dfrac{\partial E}{\partial z} =\left(\dfrac{e\left[3m_I^2 - I(I+1)\right]}{4I(2I-1)}\right)\tilde{Q}\dfrac{\partial E}{\partial z}$. | (22) |

We see that the energy shift vanishes for $I = 1/2$ as expected and depends only on the square of $m_I$ since, for example, states with $m_I \pm 1/2$ will form the same angle between the nuclear moment and the direction of the electric field gradient.

In our experiment, we will establish an electric field gradient at the nucleus by looking at absorption in a molecule where an atom of iron is surround by an asymmetric grouping of atoms. The nitroprusside ion shown in Fig. 8 has a single iron atom surrounded by five cyanyl groups (CN) and one nitrosyl group (NO). The presence of the NO breaks the symmetry and causes a strong electric field gradient at the center of the ion.

In this environment, the $I = 3/2$ excited state of ${}^{57}$Fe will be split as outlined above into two states, and we consequently observe two emission lines (see Fig. 9) separated by energy difference

$\Delta E_{quad} = \Delta E(I= 3/2, m_I = \pm 3/2) - \Delta E (I = 3/2, m_I = \pm 1/2)$. | (23) |

You will look at recoilless radiation relating to the 14.41 keV transition between the $I = 3/2$ and $I = 1/2$ nuclear states in ${}^{57}$Fe. A source of ${}^{57}$Co decays by K-electron capture to ${}^{57}$Fe in the nuclear excited state with $I = 5/2$ and then to $I = 3/2$. The 14.41 keV line is emitted when the $I = 3/2$ excited state in ${}^{57}$Fe decays to the $I = 1/2$ ground state with a lifetime of $\tau = 1.4 \times 10^{-7} \mathrm {sec}$. (See Fig. 6.)

NOTEBOOK:Based on Fig. 6, what are the expected energies of the photons emitted by ${}^{57}$Co?

Note also the mechanism for the decay from ${}^{57}$Co to ${}^{57}$Fe is *electron capture*. In this process a K-shell electron in cobalt is captured by the nucleus, thus transforming it to iron. Therefore, the decay scheme of Fig. 6 only shows *part* of the picture. The iron atom is actually unstable in the following two ways:

The nucleus of the iron is in an excited state which decays with the emission of photons. This results in the emission of the three photon energies shown above in Fig. 6. In addition, the atomic electron configuration of the iron has a vacancy in its K-shell. (This is the vacancy left when the ${}^{57}$Co nucleus *captures* an electron.) This electronic configuration is unstable and an electron from a higher shell falls into the K-shell, emitting an x-ray of an energy characteristic of the iron atom.

NOTEBOOK: Look up the K-line x-ray energy for iron here. (Note that there are several possible K-line energies since the electron can come from the $n = 2, 3,\ldots$ levels. These should all be closely spaced in energy.)

The energy of the gammas emitted from the ${}^{57}$Fe may be varied slightly (Doppler-shifted) by moving the source relative to the absorber. The source is mounted on a linear motor that is magnetically driven toward and away from the absorber. In each cycle the source experiences a constant acceleration and passes through a range of velocities of about +1 cm/sec to –1 cm/sec.

To measure the velocity of the source, one counts interference fringes from a Michelson interferometer. Fringe counting is achieved with a photo-diode, a pulse shaper, and a multi-channel scaler (MCS). The interferometer, as shown in Fig. 7, consists of a beamsplitter, a stationary mirror and a moving mirror. The moving mirror is attached to the linear motor that also moves the ${}^{57}$Co source. The interferometer is illuminated by a HeNe laser with wavelength $\lambda = 632.8\,\rm{nm}$.

As the moving mirror (and ${}^{57}$Co source) moves a distance $\delta x = \lambda / 2$, the round trip optical path length changes by $\lambda$, and a new interference fringe appears. If $n$ fringes appear, then $\Delta x = n\lambda /2$. Therefore, the velocity of the source is given by

$v = \dfrac{\Delta x}{\Delta t} = \dfrac{n\lambda}{2\Delta t}$. | (10) |

As the fringes pass across the photo-diode, they produce a varying current that is shaped into distinct pulses and fed into the pulse height analyzer (configured as a multi-channel scaler) where they are counted.

With a means of velocity calibration established, we can begin to look at the absorption of the Doppler-shifted gammas as a function of energy. To do so, we measure transmission through the absorber. Gammas that pass through the absorber are incident on a gas proportional counter. In the proportional counter, photons eject electrons from atoms in the gas. These electrons pass through the gas, ionizing other atoms, forming a cascade of electrons that are collected on a central wire, maintained at positive high voltage. The operating voltage is chosen such that the total *number* of electrons in the cascade is proportional to the *energy* of the incident photons.

The SpecTech UCS-30 (Universal Computer Spectrometer) serves as both a pulse height analyzer (PHA) and a multi-channel scaler (MCS). Specifically, the two modes (selected in software) are as follows:

**PHA - Direct In:**In*PHA*mode, the vertical axis is*number of counts*and the horizontal axis is*pulse height*(proportional to photon energy). In*Direct In*mode, we skip the spectrometer's internal amplifier (as opposed to*PHA - Pre-Amp*mode, which goes through the amplifier.)**Mössbauer - Internal:**In*Mössbauer*mode, the vertical axis is*number of counts*and the horizontal axis is*time*.

In *Mössbauer* mode, the UCS-30 is acting like a so-called multi-channel scaler (MCS). Such a device is used to display counts ($y$-axis) vs. time ($x$-axis). The *x*-axis of the MCS is divided into discrete channels, each representing a small time interval. The MCS sweeps through the channels sequentially, counting pulses for a fixed *dwell time,* $t_{dwell}$, before moving on to the next channel.
Each time the MCS reaches the last channel and returns to the first channel to start a new sweep, a pulse is sent from the MSB output on its rear panel to the Mössbauer drive unit. This pulse triggers the drive unit and starts a new cycle of motion of the ${}^{57}$Co source. Thus, after the linear motor transients have died out, MCS channels are well correlated with unique velocities.

Using the Michelson interferometer and fringe counting, one can convert the x-axis to velocities via Eq. (10). For a single pass, $\Delta t = t_{dwell}$ and $n$ is the number of fringes counted during that interval. However,$n$ can vary from pass-to-pass, and one can get a better measurement by performing many passes, allowing the data to accumulate. For a number of passes $p$, Eq. (10) is properly interpreted with *n* representing the *total* number of counts in each channel and $\Delta t$ representing the *total* time spent in that channel (i.e. $\Delta t = p * t_{dwell}$).

The apparatus should already be wired according to the connections shown in Fig. 8. (Note that *either* the amplified proportional counter output *or* the interferometer detector pulse shaper output serve as input to the PHA, depending on what we want to look at; we will never have both connected at the same time.)

Set the proportional counter high voltage to 1800 - 2000 V. Remove the plastic x-ray filter (if present), and use an oscilloscope to look at the output of the amplifier. In the midst of the other pulses you should be able to resolve two prominent bright bands of pulses and a flat-topped band of pulses clipped by the amplifier (see Fig. 9). The height of the pulse you see is proportional to the energy of the photons. Notice that by placing the x-ray filter in the beam you can decrease (or eliminate) one of the lines without drastically altering the other.

The PHA will be helpful for identifying the correct photons to count. To set up the spectrometer in PHA - Direct In mode, do the following:

- Disconnect the proportional counter output from the oscilloscope and plug it into the
*input*on the rear of the UCS-30. - Make sure the UCS-30 is connected to the computer via the USB cable and that the
*MSB*rear panel output is connected to the*BISTABLE*input on the rear of the Mössbauer drive unit. - Start the UCS-30 software. Pull down the
*MODE*menu and select*PHA - Direct In*. Drag the LLD (lower level discriminator) tab to the 5% position. Drag the ULD (upper level discriminator) tab to the 100% position. This setting allows a full range of pulse heights to be accepted, while eliminating small pulses due to amplifier noise. - Pull down the
*Settings*menu and select*High Voltage/Amplifier/ADC*. In this menu, set*Conversion Gain*to 512. This setting maps pulse heights onto channels ranging from 0 to 511.

Collect a spectrum without the x-ray filter. To do so, click on the *Start* box. Data should begin to appear. The default for the vertical axis scale is a logarithmic scale, with maximum counts = 16 million. You may change the vertical axis to be linear, with varying full-scale settings by clicking on the zoom in or out buttons.

NOTEBOOK: Recall the different energy bands that you saw on the oscilloscope. Identify the peaks in the spectrum based on their relative positions on the energy axis of the PHA.

Now place the x-ray filter between the source and detector. Collect a spectrum again.

NOTEBOOK: What effect does the x-ray filter have on the peaks in the energy spectrum? Why?

The x-ray filter has a higher cross-section for absorbing low energy photons (via the photoelectric effect) than for scatting higher energy photons (via Compton scattering) and therefore preferentially filters the 6 keV line relative to the 14.4 keV line. If the dead time on the detector is very high (greater than 10%), the filter should be used to lower the rate. If not, you can omit the filter for the remainder of the experiment.

Adjust the gain on the amplifier until the 14.4 keV line falls near the middle of the energy scale. Drag the LLD (Lower Level Discriminator) and ULD (Upper Level Discriminator) tabs to bracket the 14.4 keV peak. The discriminators will then pass only the desired gammas to be counted in the subsequent Mössbauer spectra.

NOTEBOOK: Record the lower and upper level discriminator values which bracket your peak. Save a spectrum showing both the bracketed and unbracketed peaks.

To collect a Mössbauer spectrum, pull down the Mode menu and select *Mössbauer - Internal*. (This is a special case of multi-channel scalar (MCS) mode.)

In this mode, the UCS-30 divides a single motor cycle into 512 equal-length channels. As the motor moves, the PHA counts all pulses falling in the voltage range set between the LLD and ULD during each fixed time interval and adds that to the running total in that channel. The PHA and motor remain in sync such that a particular channel on the spectrum always represents the same small interval of time in the motion of the motor.

Put another way, **the cycle is broken down into 512 pieces, each mapped onto a single channel of the spectrum. Therefore, channel in this mode can be mapped to a velocity at the point in the cycle and therefore to the Doppler energy shift at that point in the cycle.**

Pull down the Settings menu and select *Mössbauer dwell time* = $300~\mu\textrm{s}$. This sets the time for each channel to be 300 μs, which corresponds to a total cycle time of $512*300~\mu\textrm{s} = 153.6~\textrm{ms}$ (or, approximately 1/6 of a second.)

The stainless steel absorber is non-magnetic and displays the simplest Mössbauer spectrum. Select the 0.001“-thick SS 302 foil and place it between the source and the proportional counter shield. If necessary, use the x-ray filter to cut down on count rate.

While still in *Mössbauer - Internal* mode, collect a spectrum with the stainless steel sample in place. The motor will take a few cycles to come up to speed, so the first couple passes will only be noise. After a few seconds, erase the spectrum (do not stop it) to ensure the transient behavior has died out.

NOTEBOOK: Collect a Mössbauer spectrum until adequate statistics are achieved. Record the filename and the number of passes in your lab notebook. What do you expect to observe? What do you observe?

You have your first spectrum, but we still need to find the correspondence between channel and energy shift. As explained in the section on equipment, we may count fringes from the Michelson interferometer in order to calibrate our PHA time axis in units of velocity.

Disconnect the detector output from the back of the PHA and instead connect the output of the Michelson interferometer photodiode pulse shaper. (That signal should run from the pulse shaper to the tee on the speaker, then to the input on the PHA). Turn on the speaker.

Return to *Direct-In* mode. Start to collect a spectrum and **gently** wiggle the mirror back and forth by-hand. You hopefully will see a number of counts accumulating in a single channel on the spectrum as you move the mirror; this corresponds to the voltage pulses coming out of the pulse shaper as the light intensity at the photodiode changes. If you do not see an accumulation of pulses, it likely means that the pulse shaper output falls *outside* the range set by your lower and upper discriminators. Record the current positions of the upper and lower discriminators (if you haven't done so already), and then open up the range by setting both to their extreme positions. Do you see the accumulation of pulses now?

Switch back to *Mossbauer-Internal* mode. Click on the *Start* button to begin acquiring data. Erase the data after the first few passes to clear out the transient noise.

In order to get a clear and meaningful velocity calibration, we need to do a few diagnostic checks and make adjustments if necessary.

- Your first diagnostic check is to look at the spectrum which emerges. If everything is working correctly, you should see two smooth “V” shapes without hiccup – one wide and one narrow. (A little bit of curvature at the extremes is OK.)
- Your second diagnostic check is to listen to the speaker. The speaker should provide a clean “chirp” that increases in frequency and repeats about 6 times per second. (I.e., in a single second you should hear “whoop, whoop, whoop, whoop, whoop, whoop”.)
- Your final diagnostic check is to look at the DRIVE and VELOCITY signals generated by the motor on the oscilloscope. The first of these signals controls the position of the magnet which drives the motor; think of it a signal which represents the
*desired*displacement of the sample. The second signal is the*measured*velocity of the motor. The drive system uses a feedback loop to change the shape of the drive signal in order to produce a smooth linear velocity. The amount of feedback is controlled by the*fidelity*knob on the front of the motor drive.

The DRIVE signal should be quadratic and the VELOCITY signal should be a sawtooth waveform (indicating constant acceleration, or linearly increasing/decreasing velocity). Each cycle contains two regions. The longer region is the portion where you will obtain data; the shorter region is the *flyback* where the motor resets to start the main cycle again. The ideal shapes for these signals are shown in Fig. 10. If the VELOCITY and DRIVE signals differ significantly from their proper forms, you may need make adjustments to the motor. Ask the lab staff or TAs for help.

If one or more of the above diagnostic checks doesn't seem correct, adjustments may need to be made:

- If no counts are appearing on the spectrum, or if no sound is emitted by the speaker, you may need to make adjustments to the Michelson interferometer.
- If the VELOCITY output is not smooth or if the V-curve on the spectrum is not smooth, you may need to adjust the
*fidelity*knob o the drive motor.

As you make adjustments, erase – do not stop – the spectrum to clear the data and start the count over. Ask a TA or staff member for help if more than minor adjustments are needed.

NOTEBOOK:Why do we get two V shapes? What physically is going on to produce this spectrum?

NOTEBOOK:When adjustments are complete, sketch the VELOCITY and DRIVE outputs in your notebook and compare them to the sketches shown in Fig. 10.

NOTEBOOK:When the spectrum is clean, save it and note the filename in your notebook.Also, make sure to record the elapsed number of passes.

This spectrum represents the number of fringes as a function of channel. Equation (10) relates $n(ch)$ to $v(ch)$, and this can in turn be used in the Doppler relation, Eq. (4), to find the energy shift, $E'_t(ch) - E_t = \Delta E(ch) = v(ch) E_t/c$. While the raw fringe count can be fed directly into Eq. (10), it is preferable to perform a fit to the fringe data of the form $n(ch) = n_0 \left | ch - ch_0\right |$ where $n_0$ is the slope and $ch_0$ is the channel where the fringe count (and therefore velocity) goes to zero. This will allow you to smooth out any bumps or ripples that may appear in the raw data.

Also, note that while the number of fringes $n$ is always positive, the velocity changes continuously from negative (on the left branch) to positive (on the right branch). Therefore, though you fit your function with absolute value bars, you will need to relax to parenthesis (allowing both negative and positive values) when you convert from $n(ch)$ into $v(ch)$, and then ultimately to $\Delta E(ch)$. In your report, all Mössbauer spectra should be displayed with the $x$-axis converted to energy shift.

**NOTE**: For purposes of error propagation, it is usually preferable to do fits with the $x$-axis left in raw channel and to apply the energy shift calibration equation only to the extracted width and centroid fit parameters so that the calibration uncertainties can be properly propagated. In our case, the fit parameter uncertainties will be dominated by errors from the scatter of the data meaning that the calibration errors can be neglected. In this case, calibrating before or after fitting leads to the same results, so the rescaling of the $x$-axis before fit is preferred.

With the calibration set, we can now return to the stainless steel spectrum and make observations.

- If you made NO changes to the drive motor during velocity calibration, you may keep the stainless steel spectrum you previously collected.
- If you made ANY changes to the drive motor during velocity calibration, you should repeat the spectrum. Make sure to reset the lower and upper discriminators in
*PHA - Direct In*mode if needed.

The energy of the gammas emitted by the source have a characteristic distribution known as the Breit-Wigner or Lorentzian distribution (shown in Fig. 11). Mathematically, this distribution has the form

$I(E) = I_0 \dfrac{\left(\Gamma/2\right)^2}{\left(E - E_0\right)^2 + \left(\Gamma/2\right)^2}$, | (11) |

where $E_0$ is the transition energy (i.e, the center of the peak), $I_0$ is the peak intensity, and $\Gamma$ is the natural linewidth of the transition (i.e, the full-width at half maximum (FWHM) of the peak)

Note, however, that in our experiment we do not see the emission directly, but instead measure a *dip* in the transmission spectrum due to absorption. The probability for absorption will be the convolution of the absorption spectrum (a Lorentizan with FWHM $\Gamma$) and the emission spectrum (a Lorentizan with FWHM $\Gamma$), such that the measured dip is a Lorentizian with FWHM $2\Gamma$. Therefore, the observed linewidth is related to the true linewidth as

$\Gamma_{obs} \approx 2\Gamma$. | (12) |

We use an approximately equal sign in Eq. (12) because the finite thickness of the absorber can lead to additional widening and distortion effects. Therefore, $\Gamma_{\textrm{obs}} = 2\Gamma$ only in the limit of zero thickness.

ANALYSIS: Determine the position of the center of the single dip (with uncertainty). Is the dip located at the zero velocity channel? If not, this is evidence of the isomer shift. Quantify the magnitude of this shift.

ANALYSIS:Measure the full width at half maximum (FWHM, with uncertainty) of the absorption dip. Using the correction formula of Eq. (12), determine the lifetime of the $I = 3/2$ state. How well does the value compare with the literature value of $141.4 \pm 1.4 \textrm{ns}$ [13]? The stainless steel sample is fairly thick, which distorts the lineshape and widens the linewidth. Is your result consistent with this known issue?

To observe the Zeeman effect, use an absorber of iron, enriched in the isotope 57Fe (to make the effect more prominent). Because iron is ferromagnetic, it will have a slight magnetization.

NOTEBOOK: Obtain a Mössbauer spectrum for the iron absorber and do another velocity spectrum afterward (to check the calibration). Save both and record the filenames in your notebook. Make note of the locations of the dips and verify that you observe as many features as you expect. (If the second calibration is consistent with the first one, you may use the first one for both sets of data.)

ANALYSIS:Measure the FWHM (with uncertainty) of one of the absorption dips. Again, using Eq. (12), determine the lifetime of the $I = 3/2$ state and compare it both to the result above for stainless steel and to the literature value. You should find that this dip is narrower that the stainless steel dip. What does this mean about the effective thickness of the absorber sample?

ANALYSIS: Explain the origin of each dip in terms of the allowed Zeeman transitions. Measure the separations between the lines (with uncertainties) and determine the energy differences $\Delta E_0$ and $\Delta E_1$ between the Zeeman-split levels of the $I = 1/2$ and $I = 3/2$ states.

ANALYSIS:From the above energy difference determinations, do the following:

(1) Calculate the ratio $\left|\frac{u_0}{u_1}\right|$ of the magnetic moments and, using the literature value of $\mu_0 = (0.09024 \pm 0.00007)\mu_N$ [11], determine $\mu_1$.

(2) Determine the magnitude of the magnetic field (the $H$-field) at the location of the iron nucleus.

Experimentally-determined values for comparison can be found in Ref. [7]. Within this work, the symbol “nm” is used to represent the nuclear magneton, $\mu_N$, and the H-field is given in oersteds, the CGS unit for a magnetic field inside a material. You can assume that 1 oersted is $10^{-4}$ Tesla. Additionally, a value for the magnetic moment ratio can be found in Ref. [14].

*Part II of the experiment is intentionally left open-ended. You may need to consult outside resources for more theory information or equipment manuals for details about the apparatus. Given your experience collecting data in Part I, use your judgement to determine effective collection and analysis strategies, and budget your remaining time in lab appropriately.*

*Below is a list of possible topics to explore. Based on your interests and the quality of your work after Part I, you and the faculty instructor may discuss alternate goals. You are encouraged to bring ideas and propose experiments at your meeting, but you should also be prepared to defend approaches which stray far from the topics outlined below. *

From a stainless steel Mössbauer spectrum, measure the depth of the absorption dip. The fractional absorption is defined by

$A = \frac{N_{\infty} − N_{dip}}{N_{\infty} − N_{bg}}$ | (25) |

where $N_{\infty}$is the number of counts per channel far from the dip, $N_{dip} is the number of counts at the bottom of the dip, and N_{bg} is the number of the N_{\infty} counts that are due to background.

**QUESTION**: Calculate the fractional absorption. Investigate the theory behind recoilless absorption, and determine the effective absorber thickness, $T_{A}$, and the recoilless fraction, $f$. How well does this measured $f$ agree with the predicted value?

From a classical perspective, it seems rather improbable that a single nucleus in a crystal lattice could cause the entire lattice to recoil when it emits (or absorbs) a photon. In the classical picture, we view the lattice as a large number of atoms attached to each other by springs. If a photon comes into this lattice and is absorbed by a single nucleus, we would expect that this absorbing nucleus would recoil relative to the other atoms. This recoil would then set up vibrational waves in the lattice, which would propagate through the solid. It would seem impossible for a nucleus in a classical crystal to absorb a photon without recoiling.

When we treat the crystal quantum mechanically, however, there is an important difference in how we view the crystal. Each vibrational mode of the crystal is now treated as an independent harmonic oscillator, with a characteristic frequency of oscillation $\omega_i$. These quantized modes are called **phonons**. Importantly, these modes can only be excited in discrete energy steps; a phonon with frequency $\omega_i$ must always have energy $n\hbar\omega_i$ relative to the ground state, where $n$ is an integer. (Note the similarity between this quantization and the energy levels of the conventional harmonic oscillator.) In the classical picture, we could give a vibrational mode of the crystal an arbitrarily low energy, and thus many different vibrational modes could be excited by the movement of a single nucleus. However, in the quantum picture, many vibrational modes will remain unexcited, since the energy that they would receive classically is smaller than $\hbar\omega_i$ (the minimum excited energy they can have).

In general, then, some fraction $f$ of the emissions will leave the crystal in its ground state, without exciting any phonons at all. Remarkably, we can actually say quite a bit about what this fraction will be, without having detailed knowledge of the phonon energies in the system. We will give a brief calculation of $f$ here; more details can be found in [1].

Generally, the transition from an initial quantum state $\left|\, \psi_{i} \right>$ to a final state $\left|\, \psi_f \right>$ under the influence of some external stimulus will be determined by the matrix element of some operator $\mathcal{T}$:

$\mathcal{M}_{i\rightarrow f} = \left< \psi_f \right | \mathcal{T} \left| \psi_{i} \right>$. | (5) |

In our case, the states $\left|\, \psi_{i} \right>$ and $\left|\, \psi_{f} \right>$ will be a direct product between a vibrational state of the lattice (characterized by a set of integers $n$ giving the “level” to which each phonon mode $i$ is excited) and some set of variables describing the state of the absorbing or emitting nucleus. We will denote those initial and final states by $\left|\, n_{i},\phi_i \right>$ and $\left|\, n_{f},\phi_f \right>$ respectively.

The operator $\mathcal{T}$ will depend on the position of the individual nucleons, their momenta, and their spins, as well as the photon's momentum $\mathbf{K}$. However, it can be shown [1, 2] that this operator must, in general, take the form

$\mathcal{T} = \exp \left( \frac{i \mathbf{K} \cdot \mathbf{X}}{\hbar}\right) U$. | (6) |

where $\mathbf{X}$ is the center of mass of the nucleus and the operator $U$ depends only on the internal degrees of freedom of the nucleus. This implies that our matrix element $\mathcal{M}$ can be split up into two factors, one depending on the transitions of the lattice and the other depending on the transitions of the nucleus:

$\mathcal{M}_{i\rightarrow f} = \left<\, n_{i\vphantom{f}}\right| \exp \left( \frac{i \mathbf{K} \cdot \mathbf{X}}{\hbar}\right)\left|\, n_{i\vphantom{f}}\right> \cdot \left< \phi_f \right| U \left| \phi_{i\vphantom{v}}\right>$. | (7) |

The probability of a transition from $\left|\, n_{i\vphantom{f}},\phi_i \right>$ to $\left|\, n_{f},\phi_f \right>$ will then be proportional to $\left|\mathcal{M}\right|^2$. The exact transition rate will depend on the matrix element $\left< \phi_f \right| U \left| \phi_{i\vphantom{v}}\right>$, but if all we are concerned with is the relative probabilities of the lattice transitions for a given nuclear transition, then this factor will be the same for each lattice transition and will cancel out. This leaves the relative probability of each lattice transition as

$\mathcal{P}_{n_i\rightarrow n_f} = \left|\left<\, n_{i\vphantom{f}}\right| \exp \left( \frac{i \mathbf{K} \cdot \mathbf{X}}{\hbar}\right)\left|\, n_{i\vphantom{f}}\right>\right|^2$. | (8) |

In particular, the probability that an emission will take place which leaves the lattice in the state $\left|\, n_{i\vphantom{f}}\right>$ – which is termed the **recoilless fraction** – will be

$f = \mathcal{P}_{n_i\rightarrow n_f} = \left|\left<\, n_{i\vphantom{f}}\right| \exp \left( \frac{i \mathbf{K} \cdot \mathbf{X}}{\hbar}\right)\left|\, n_{i\vphantom{f}}\right>\right|^2 = \left| d^3\mathbf{X}\psi_i^*(\mathbf{X} )\psi_i(\mathbf{X} )\exp \left( \frac{i \mathbf{K} \cdot \mathbf{X}}{\hbar}\right)\right|^2$ | (9) |

where $\psi_i(\mathbf{X})$ is the wavefunction of the state $\left|\, n_{i\vphantom{f}}\right>$.

In our case, the initial state of the lattice will not be a pure state but rather a mixed thermal state. In this case, the above formula for the recoilless fraction still holds, but with $\psi_i^*(\mathbf{X} )\psi_i(\mathbf{X} )$ replaced by the averaged probability density $p(\mathbf{X})$,

$f = \left| \,d^3\mathbf{X}\,p(\mathbf{X})\exp \left( \frac{i \mathbf{K} \cdot \mathbf{X}}{\hbar}\right)\right|^2$. | (10) |

This probability density will, to a very good approximation, be a Gaussian with some mean-squared displacement $\xi^2$,

$p(\mathbf{X}) = \dfrac{1}{\left(2\pi \right)^{3/2}\xi^3} exp \left[-\dfrac{\mathbf{X}^2}{2\xi^2}\right]$ | (11) |

Performing the integral then yields the rather simple answer

$f = exp \left[-\dfrac{K^2\xi^2}{2\hbar^2}\right] = exp \left[-\dfrac{2\pi^2\xi^2}{\lambda^2}\right]$, |

where $\lambda$ is the wavelength of the emitted photon.

We see from the above formula that the recoilless fraction only depends on the mean-squared displacement of the nuclei in a thermal state; we don't need to know any other information about the dynamics in the crystal. In general, we will only see significant recoilless absorption (and emission) if the argument of the exponential in Eq. (12) is less than 1. This implies that we should see significant recoilless fraction when $\sqrt{\xi^2/\lambda^2}$ is less than 0.1. Intuitively, this answer makes sense; the phonon modes will be more likely to remain unexcited if the momentum $\mathbf K$ they receive is sufficiently small – the harder we “hit” a nucleus, the more likely we are to give a phonon mode its minimum excitation energy – or if the nuclei are sufficiently tightly bound – i.e., $\sigma^2$ is small (since this will increase the minimum excitation energy of the phonons.)

QUESTION: For iron metal, $\xi^2$ has been measured via neutron diffraction to be 0.0021 ± 0.0003 Å^2 at room temperature [13]. What is the recoilless fraction f of the 14.4 keV photon emission for this situation? (Be careful with your factors of $2\pi$!)

We have been assuming that the energy of the lattice is constant throughout the recoilless absorption process. This is, however, not strictly true. Classically, the energy of the lattice is given by

$U = \sum\limits_{i}\dfrac{\mathbf{p}_i^2}{2m_i} + V(\mathbf{x}_i,\, \dots,\mathbf{x}_N )$. | (26) |

Even though the recoilless absorption causes no change in the positions $\mathbf{x}_i$ of the nuclei and negligible change in their momenta $\mathbf{p}_i$, there is still a change to the mass $m_e$ of the absorbing nucleus. The absorption energy $E_t$ is equivalent to an additional mass (according to Einstein), and thus when the absorbing nucleus gains this energy its mass will increase by $\delta m_e = E_t/c$. The expected change to the lattice energy will then be given by

$\delta U = -\dfrac{\delta m_e}{m_e}\left<\dfrac{\mathbf{p}^2_e}{2m_2}\right> = -E_t \dfrac{\left<\mathbf{v}^2_e\right>}{2c^2}$. | (27) |

By conservation of energy, then, the absorption energy will be raised by $-\delta U$:

$\Delta E_{therm} = -\delta U = E_t \dfrac{\left<\mathbf{v}^2_e\right>}{2c^2}$. | (28) |

This implies that if $\left<\mathbf{v}^2_e\right>$ is different for source and absorber, the center of the absorption peak will not occur at zero velocity. We can adjust the value of $\left<\mathbf{v}^2_e\right>$ for our absorber by heating it (while leaving the source at room temperature); this will cause the thermal motion of the atoms to increase, thereby shifting the peak absorption energy relative to the emission energy by $\Delta E_{therm}$.

To make quantitative predictions concerning the magnitude of this shift, we need to have some model for the vibrational dynamics of the solid. Specifically, if we know the density of phonon states $g(\omega)$, then we can show that [12]

$\left<\mathbf{v}^2_e\right>=\dfrac{3\hbar}{m}\displaystyle\int d\omega \coth \left( \dfrac{\hbar \omega}{2k_BT}\right)g(\omega )\omega$ | (29) |

where $m$ is the mass of the nucleus. The simplest realistic model for $g(\omega)$ is the **Debye model**, where $g(\omega)$ is assumed to be quadratic up to a cutoff frequency $\omega_D$ and vanishes above $\omega_D$. Under this assumption, we can rewrite Eq. (29) in terms of the Debye temperature $\Theta_D = \hbar\omega_D/k_B$ (where $k_B$ is the Boltzmann constant) as

${\color{white}{.}} \left<\mathbf{v}^2_e\right>= \dfrac{9k_B\Theta_D}{m}\left[\dfrac{1}{8} + \left( \dfrac{T}{\Theta_D}\right)^4 \displaystyle\int_0^{\Theta_D/T} dx \dfrac{x^3}{e^x -1}\right]$ | |

${\color{white}{.}} \hphantom{\left<\mathbf{v}^2_e\right>} \approx \dfrac{3k_BT}{m}\left[1+ \dfrac{1}{20}\left( \dfrac{\Theta_D}{T}\right)^2 -\dfrac{1}{1680}\left( \dfrac{\Theta_D}{T}\right)^4 + \dfrac{1}{90720}\left( \dfrac{\Theta_D}{T}\right)^6 + \dots \right]$ | (30) |

for $T > \Theta_D/2$. Note that in the limit of high temperature, we have $\left<\mathbf{v}^2_e\right> \approx \frac{3k_BT}{m}$ as we would expect from the equipartition theorem.

We can also relate the temperature shift to the heat capacity of the solid. Assuming that the lattice forces are harmonic, the total kinetic energy of the vibrating atoms $T=Nm\left<\mathbf{v}^2_e\right>/2$ will be one-half of the total lattice energy $U$. This implies that

$\dfrac{\Delta E_{therm}}{E} = \dfrac{T}{nmc^2} = \dfrac{U}{2Nmc^2}$ | (31) |

and differentiating with respect to $T$ at constant $P$, we conclude that

$\dfrac{\partial}{\partial T}\left. \left(\dfrac{\Delta E_{therm}}{E}\right)\right|_P = \dfrac{C_P}{2Nmc^2}$ | (32) |

where $C_P$ is the heat capacity per mass at constant pressure of the absorber.

Raising the temperature of the absorber will also cause the mean-squared displacement of the nuclei to increase; this implies, recalling the theory above, that the recoilless fraction $f$ will decrease as $T$ increases. Specifically, it can be shown that under the Debye model,

${\color{white}{.}} \left<\mathbf{x}^2\right>= \dfrac{3\hbar^2}{4mk_B\Theta_D}\left[1+4\dfrac{T}{\Theta_D}^2 \displaystyle\int_0^{\Theta_D} dx \dfrac{x}{e^x -1}\right]$ | |

${\color{white}{.}} \hphantom{\left<\mathbf{x}^2\right>}\approx \dfrac{3\hbar^2T}{mk_B\Theta_D^2}\left[1+\dfrac{1}{36}\dfrac{T}{\Theta_D}^2 - \dfrac{1}{3600}\left(\dfrac{T}{\Theta_D}\right)^4+\,\dots \right]$. | (33) |

This variation in $\left<\mathbf{x}^2\right>= \xi^2$ can then be applied to our formula for the recoilless fraction – Eq. (12) – and compared to experimental results.

QUESTION: Look up the Debye temperature for Iron. Is the approximate formula for the mean-squared velocity in Eq. (33) okay for our situation?

Place the stainless steel absorber in the heater and raise its temperature to 100 $^{\circ}$C or 200 $^{\circ}$C. Obtain Mössbauer spectra and velocity calibration spectra. Observe the shift of the central peak.

QUESTION: Is it in the correct direction? How large is it? Is the size of the shift in agreement with the predictions of the second-order Doppler treatment? From the size of the shift calculate the lattice specific heat of iron.

The Debye theory makes predictions about the change in fractional absorption with temperature. Because the mean square displacement $\left<\mathbf{x}^2\right>$ depends in a specific way on temperature, you can calculate the change in the recoilless fraction of the absorber when it is heated. The recoilless fraction of the source does not change, so you can predict how much the observed absorption should change. Measure the absorption as in the previous section and compare with expectations.

To observe quadrupole splitting you need an absorber containing iron in a non-symmetric environment. In such a case, the nuclear quadrupole moment interacts with the electric field gradient, splitting the central transition line in two.

NOTEBOOK: Place the sodium nitroprusside absorber, $\textrm{Na}_2[\textrm{Fe}(\textrm{CN})_5 \textrm{NO}]\cdot 2\textrm{H}_2\textrm{O}$, in the beam and collect a Mössbauer spectrum and another velocity spectrum. Save both and record the filenames in your notebook. Note the locations of the dips.

REPORT: From Eq. (13), show that you expect two dips symmetric about the non-split energy and measure the difference in energies between the two observed features (with uncertainties). By substituting the appropriate quantum numbers into Eq. (13), you can compare the measured energy difference to the prediction and determine $Q dE/dz$, the product of the quadrupole moment for the $I = 3/2$ state and the field gradient. Given that $Q$ is typically on the order of 0.1 - 1.0 barn (1 barn = $10^{-24}$ cm${}^2$), how large does this make the field gradient? Can you make a plausibility argument to show that this field gradient value is reasonable given the scales of the molecule involved?

Obtain a Mössbauer spectrum of Metglas (a metallic glass) at room temperature and at various temperatures up to and beyond its Curie point. Note how the temperature affects the Zeeman splitting.

Can you raise temperature above the glass' melting point? If you can maintain such a temperature, obtain spectra above the melting point and back at room temperature after it has cooled.

This section has scans of printed materials that have been used in this lab in the past