5 Mössbauer Spectroscopy with High Spatial Resolution …
241
where σ 0 is the resonance cross section of the transition (= 2.56 × 10
–18 cm
2
for the 14.4 keV
57 Fe transition), f a is the recoil-free fraction of the sample, and
n a is the number of
57 Fe atoms per cm
2 in the sample. Equation 5.6 is the equivalent
of Eq. 5.1 for the sample. In the same way that the thin source limit is defined as
t s < < 1, the thin absorber approximation is valid for t a < < 1. However while the ideal
value for t s is zero, the value of t a must be non-zero in order to obtain a signal. There
are analytical advantages to values of t a within the thin absorber approximation [12],
but such values are rarely the best choice for optimising the quality of the spectrum
[45]. Extensive discussions of how to choose t a are given in [1, 4].
A trade-off occurs when sample thickness is increased: a greater thickness means
more iron and hence a stronger signal, but it also increases the amount of other
elements in the sample which attenuates the signal due to electronic absorption
and scattering. The latter is quantified through the total sample mass absorption
coefficient which is given by:
μ =
i
f i μ i
(5.7)
where f i is the mass fraction of the ith element and μ i is the mass absorption coefficient of the ith element at 14.4 keV. The mass absorption coefficients for all elements
are listed for selected energies in [46]. If sample thickness is expressed as an area
density t
(for example in units of g/cm
2 ), the maximum signal to noise ratio for a
sample as a function of t
(optimum thickness) occurs at t
= 2/μ when non-resonant
background is low (i.e., the sample contains no heavy elements) and t
= 1/μ when
non-resonant background is high [1, 46].
These relations can be applied to Fe 0.4 Mg 0.6 Si 0.63 Al 0.37 O 3 , whose spectra are
shown in Fig. 5.4. The total sample mass absorption coefficient calculated from
Eq. 5.7 is 17.3 cm
2 /g for 14.4 keV radiation, which means the optimum thickness
t
is 115.9 mg/cm
2 of sample. The physical thickness can be calculated from 10 t
/ρ
where ρ is the density of the sample. For the density 5 g/cm
3 , the physical thickness is
232 μm, which corresponds to a dimensionless effective thickness of 9.3 according
to Eq. 5.6 for iron with natural isotopic abundance (2.14%
57 Fe). However the sample
used in Fig. 5.4 was constrained to a physical thickness of 15 μm because of diamond
anvil cell geometry, which would have been a dimensionless effective thickness of 0.6
for natural
57 Fe abundance. To obtain a stronger signal, the sample was synthesised
with 90% enriched
57 Fe, giving a dimensionless effective thickness of 25.4 according
to Eq. 5.6. While this value is high and causes substantial thickness effects, a strong
signal is required to overcome the loss of intensity due to absorption by the diamonds.
In addition to electronic absorption by elements within the sample, signal will be
lost due to surrounding material that γ-rays pass through. Examples include substrates
that the sample is mounted on, liquid if the sample is suspended in solution, and chambers and/or windows associated with samples in extreme environments (for example
a diamond anvil cell) . The intensity of transmitted radiation can be calculated from:
241
where σ 0 is the resonance cross section of the transition (= 2.56 × 10
–18 cm
2
for the 14.4 keV
57 Fe transition), f a is the recoil-free fraction of the sample, and
n a is the number of
57 Fe atoms per cm
2 in the sample. Equation 5.6 is the equivalent
of Eq. 5.1 for the sample. In the same way that the thin source limit is defined as
t s < < 1, the thin absorber approximation is valid for t a < < 1. However while the ideal
value for t s is zero, the value of t a must be non-zero in order to obtain a signal. There
are analytical advantages to values of t a within the thin absorber approximation [12],
but such values are rarely the best choice for optimising the quality of the spectrum
[45]. Extensive discussions of how to choose t a are given in [1, 4].
A trade-off occurs when sample thickness is increased: a greater thickness means
more iron and hence a stronger signal, but it also increases the amount of other
elements in the sample which attenuates the signal due to electronic absorption
and scattering. The latter is quantified through the total sample mass absorption
coefficient which is given by:
μ =
i
f i μ i
(5.7)
where f i is the mass fraction of the ith element and μ i is the mass absorption coefficient of the ith element at 14.4 keV. The mass absorption coefficients for all elements
are listed for selected energies in [46]. If sample thickness is expressed as an area
density t
(for example in units of g/cm
2 ), the maximum signal to noise ratio for a
sample as a function of t
(optimum thickness) occurs at t
= 2/μ when non-resonant
background is low (i.e., the sample contains no heavy elements) and t
= 1/μ when
non-resonant background is high [1, 46].
These relations can be applied to Fe 0.4 Mg 0.6 Si 0.63 Al 0.37 O 3 , whose spectra are
shown in Fig. 5.4. The total sample mass absorption coefficient calculated from
Eq. 5.7 is 17.3 cm
2 /g for 14.4 keV radiation, which means the optimum thickness
t
is 115.9 mg/cm
2 of sample. The physical thickness can be calculated from 10 t
/ρ
where ρ is the density of the sample. For the density 5 g/cm
3 , the physical thickness is
232 μm, which corresponds to a dimensionless effective thickness of 9.3 according
to Eq. 5.6 for iron with natural isotopic abundance (2.14%
57 Fe). However the sample
used in Fig. 5.4 was constrained to a physical thickness of 15 μm because of diamond
anvil cell geometry, which would have been a dimensionless effective thickness of 0.6
for natural
57 Fe abundance. To obtain a stronger signal, the sample was synthesised
with 90% enriched
57 Fe, giving a dimensionless effective thickness of 25.4 according
to Eq. 5.6. While this value is high and causes substantial thickness effects, a strong
signal is required to overcome the loss of intensity due to absorption by the diamonds.
In addition to electronic absorption by elements within the sample, signal will be
lost due to surrounding material that γ-rays pass through. Examples include substrates
that the sample is mounted on, liquid if the sample is suspended in solution, and chambers and/or windows associated with samples in extreme environments (for example
a diamond anvil cell) . The intensity of transmitted radiation can be calculated from: