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To calculation of particle number, the proportion of metal elements on the receiving plate is measured by EDS. After ejection, the ratio of the metal elements reaching the receiving plate to the total number of metal elements ejected is calculated. The calculation is divided into two parts: (1) the number of particles of metal elements in the damaged area is determined, and (2) the number of particles of metal elements on the receiving plate is computed.
The number of particles of metal films ejected is N1, which is expressed as:
$$ {{N}}_{1}=\frac{Ad\rho }{M}\cdot NA $$ (1) where
$ A $ is the damaged area after eruption on the surface of the metal films;$ d $ is metal film thickness;$ \;\rho $ is metal density; M is the molar mass of the metal; and$ NA= $ $ 6.02\times {10}^{23} $ .The distribution area of the particles on the receiving plate is a circle. Starting from the center of the circle, along a certain direction of the radius, r is a fixed distance for the EDS test and the test areas S are all the same until the metal elements no longer appear. The result of the EDS test is the proportion of metal elements in several different test areas, which is
$ {{P}}_{0}{\text{%}}$ ,$ {{P}}_{1}{\text{%}}$ ,$ {{P}}_{2}{\text{%}}, \cdots, {{P}}_{{n}}{\text{%}} $ . Furthermore, the number of particles in the ring area with a width of ∆r=r between the test points is the same, and this number is an arithmetic average of the density of the inner and outer rings.The number of metal particles received on the receiving plate is N2, which is expressed as:
$$ {{N}}_{2}=\sum _{n=0}^{m}{A}_{n}{\rho }_{n} $$ (2) where
$ {A}_{n} $ is the nth ring area.$ {A}_{0} $ is a circle with a radius of r and the center of the circle as the circle center,$ {A}_{0}={{\pi }}{r}^{2}$ ,${A}_{1}=3{{\pi }}{r}^{2}$ ,${A}_{2}=5{{\pi }}{r}^{2}, \cdots, {A}_{n}=(2n+ 1){{\pi }}{r}^{2} $ . The density of particles in the nth ring area is$ {\rho }_{n} $ :$$ {\rho }_{n}=\dfrac{{N}^{'}\dfrac{{P}_{n}{\text{%}}+{P}_{n+1}{\text{%}}}{2}}{S}=\dfrac{3h{\rho }_{\rm {SiO}_{2}}NA}{{M}_{\rm {SiO}_{2}}}\cdot \dfrac{{P}_{n}{\text{%}}+{P}_{n+1}{\text{%}}}{2} $$ (3) where
$\;{N}^{'} $ is the total number of SiO2 particles in the test area S and detection depth$ h $ . Nʹ is approximated to the total number of particles.Finally, the percentage of particles of metal elements received by the receiving plate
$ {\rm{\eta }} $ is expressed as:$$ {\rm{\eta }}=\frac{{N}_{2}}{{N}_{1}}\times 100{\text{%}} $$ (4) Then, the experimental results are calibrated by a series of standard samples. Using the aforementioned method to calculate the density of the number of particles of metal elements in the EDS test area, the number of particles in area S and depth h is approximately considered as the total number of test particles in the volume. However, the number of particles that EDS can detect plummets with the penetration depth. Therefore, the result needs to be corrected and calibrated. The steps for producing the standard sample are as follows:
(1) An Al film layer is coated with a thickness of d0 (1 µm) on a fused silica substrate, marked as Class I sample;
(2) SiO2 film layers of different thicknesses are coated onto Class I sample; the thicknesses of the SiO2 film layers are integral multiples of 0.25 µm, marked as Class II samples;
(3) The Al ratio of Class I and Class II samples are measured by EDS, and the Al ratio of Class II samples are modified by using the measurement results of Class I sample;
(4) The thickness of the SiO2 film layer and the modified Al ratio curve are plotted. When the Al ratio is 0, the thickness of the SiO2 film layer is the penetration depth of EDS.
As shown in Fig.2, the EDS penetration depth is approximately 1.65 μm, and the total number of particles detected in Equation (3),
${N}{'}=\dfrac{3Sh{\rho }_{\rm {SiO}_{2}}NA}{{M}_{\rm {SiO}_{2}}}$ is fixed:Figure 2. EDS measurements of 1 µm Al layer embedded in different depths of SiO2 film. The horizontal coordinate is the thickness of the SiO2 film layer, and the ordinate is the proportion of the quality of Al elements obtained by EDS. The measuring area is 50 μm×100 μm, the scanning time is 25 s, and the electron beam energy is 10 keV
$$ \begin{split} {N}{'}=&\dfrac{3S{\rho }_{\rm {SiO}_{2}}NA}{{M}_{\rm {SiO}_{2}}}{\int }_{0}^{1.65}q{\text{%}}\cdot {\rm d}h\approx\\ & \dfrac{3S{\rho }_{\rm {SiO}_{2}}NA}{{M}_{\rm {SiO}_{2}}}\sum _{n=1}^{7}0.25\times \dfrac{{q}_{n}{\text{%}}+{q}_{n+1}{\text{%}}}{2} \end{split} $$ (5) where
$ {q}_{n}\text{%} $ is the proportion of particles that can be detected by EDS under different thicknesses obtained from the experiments; and 0.25 μm is the film thickness step.Equation (3) becomes:
$$ {\rho }_{n}=\dfrac{{N}{'}\dfrac{{p}_{n}{\text{%}}+{P}_{n+1}{\text{%}}}{2}}{S}=\dfrac{3{h}^{'}{\rho }_{\rm {SiO}_{2}}NA}{{M}_{{\rm SiO}_{2}}}\cdot \dfrac{{p}_{n}{\text{%}}+{P}_{n+1}{\text{%}}}{2} $$ (6) where
$ {h}{'}=\displaystyle\sum _{n=1}^{7}0.25\times \dfrac{{q}_{n}{\text{%}}+{q}_{n+1}{\text{%}}}{2} $ can be defined as the equivalent penetration depth. The result of this calculation is$ {h}{'}=0.27\;{\text{μ}} {\rm{m}} $ . Actually, the results of EDS measurements of several metals are compared. The difference between the penetration depth and equivalent depth is not significant, which means that the penetration ability of specific X-ray signal to SiO2 or fused silica is similar.Under very high laser fluence, the metal layer and substrate of the Class I sample are damaged simultaneously, and most of the ejected particles are microns in size. The main component is SiO2, which is mainly from the substrate itself, and the metal elements adhere to the SiO2 particles. By contrast, under proper laser fluence, only the metal layer is damaged, and thermal evaporation ejects additional particles in atomic and ionic states. After a period of ejection and flight, the particles are deposited on the surface of the receiving plate. Through EDS measurements, the relative mass ratio of the metal elements can be obtained. The experimental results show that most of the metal elements are distributed within 2 mm on the receiving plate center which is placed 40 mm behind the sample, and the size of the film-peeling area is about 100–200 μm by 0.1 mm laser spot diameter.
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Differences in environments and laser fluences. Using Equations (1), (2), (4) and (6), the total number of atoms ejected by the Al film layer, the total number of atoms deposited on the receiving plate, and the deposited proportion can be determined. The distribution characteristics of Al elements on the receiving plate under different laser fluences in the atmospheric and vacuum environments are compared. As shown in Fig.3 and Tab.1, the thickness of Al film is 1 μm, which is coated by electron beam evaporation. The total number of ejected particles, the number of deposited particles, and the deposited proportion are roughly calculated. Figure 4 gives the damage morphology of Al film. Because of the stable film structure and the small internal stress, the damage morphology shows to be symmetrical.
Figure 3. Distribution characteristics of Al particles in atmospheric and vacuum environments under different laser fluences
Table 1. Number of Al particles received by the receiving plate under different conditions
Environment Laser fluence/J·cm−2 Total ejected particles Total received particles Percentage of reception Atmosphere 33 1.2481×1016 7.2583×1015 58.16% 42 1.3814×1016 7.2847×1015 52.73% 50 1.9798×1016 1.0892×1016 55.01% Vacuum 33 1.2011×1016 4.8924×1015 40.73% 42 1.4448×1016 5.6541×1015 39.13% 50 2.0198×1016 7.3754×1015 36.52% Under the same laser fluences, the size of the damage of Al films in the atmospheric environment is close to that of the vacuum environment. In the atmospheric environment, about 50%-60% of the metal particles are usually received by the receiving plate. In the vacuum environment, this proportion is slightly lower. Because the oxygen-assisted combustion and reinforcement of plasma mass will not occur in the process of metal film damage, in which the plasma eruption will exert a secondary heating and kinetic energy transfer impact on metal atoms. Figure 5 shows the transient images of Al film in vacuum and atmospheric environment. The light of the detection area in the transient images is shielded, and a black region is formed. The transient eruption profile of the metal is related to the melting point, heat transfer coefficient, local temperature, flammability, and level of thermal evaporation. By comparison, it is easier to form strong plasma region in atmospheric environment. Moreover, in the atmospheric environment, the blocking effect of atmospheric molecules on the ejected atoms during flight is not significant.
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Compared with Al film, W film has a higher melting point (Al has a melting point of 660 ℃ and W has a melting point of 3410 ℃). The eruption process caused by thermal damage is affected by the high melting point and heat transfer of W film, which is significantly different from Al. In particular, the thermal stress required to cause damage of the same size and thermal evaporation effect is greater, and the laser energy is stronger.
Figure 6 shows the ejection transient images of 5 µm thick Al and W films in a vacuum environment at 90 ns delay. The ejection morphology is related to the metal characteristics, and the difference is very obvious. These characteristics also determine the distribution and proportion of the ejected atomic metal elements, as shown in Tab.2 and Fig.7. The receiving plate receives W significantly less than Al.
Table 2. Number of Al and W particles received by the receiving plate under different conditions
Environment Fluence/J·cm−2 Element Total ejected particles Total received particles Percentage of reception Atmosphere 33 Al 1.39428×1016 9.04330008×1015 64.86% W 1.09506×1016 2.28100998×1015 20.83% Vacuum 33 Al 1.34463×1016 7.80557715×1015 58.05% W 8.71092×1015 1.965183552×1015 22.56%
Distribution characteristics of metal film eruption induced by nanosecond pulse laser
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摘要: 大功率纳秒脉冲激光辐照金属膜层时会发生热力损伤,产生的高温和高压会引起金属膜层的热蒸发,从而向外喷射颗粒,大多数喷射粒子处于原子和离子状态。在能量色散光谱分析测试中,文中使用多组标准样品进行比较实验,校准测试结果,并给出了一种根据结果计算沉积原子数的估计方法。此外,在此基础上比较了不同真空度的真空环境和大气环境下铝膜喷发的差异,并且对比了不同熔点的金属膜的喷发及空间分布特征。通过结合泵探测技术捕获的瞬态图像,进一步分析和解释了实验结果。Abstract: Thermo-mechanical damage occurs when a metal film layer is irradiated by a high-power nanosecond pulse laser. High temperature and high pressure induce the thermal evaporation of the metal film layer and particles outward ejection. Most of the ejected particles are in atomic and ionic states. In the energy dispersive spectroscopy analysis test, multiple sets of standard samples for comparison experiments was used to calibrate the test results in this paper, and an estimated method was given to calculate the number of deposited atoms based on the results. The differences of the eruption of Al film under different vacuum environments and atmospheric environments irradiated by different laser fluences were compared. And the eruption and spatial distribution characteristics of metal films with different melting points were compared. The experimental results were further analyzed and explained by combining the transient images captured by pump-probe technology.
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Key words:
- nanosecond pulse laser /
- metal film /
- vacuum /
- evaporation and ejection of materials
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Figure 2. EDS measurements of 1 µm Al layer embedded in different depths of SiO2 film. The horizontal coordinate is the thickness of the SiO2 film layer, and the ordinate is the proportion of the quality of Al elements obtained by EDS. The measuring area is 50 μm×100 μm, the scanning time is 25 s, and the electron beam energy is 10 keV
Table 1. Number of Al particles received by the receiving plate under different conditions
Environment Laser fluence/J·cm−2 Total ejected particles Total received particles Percentage of reception Atmosphere 33 1.2481×1016 7.2583×1015 58.16% 42 1.3814×1016 7.2847×1015 52.73% 50 1.9798×1016 1.0892×1016 55.01% Vacuum 33 1.2011×1016 4.8924×1015 40.73% 42 1.4448×1016 5.6541×1015 39.13% 50 2.0198×1016 7.3754×1015 36.52% Table 2. Number of Al and W particles received by the receiving plate under different conditions
Environment Fluence/J·cm−2 Element Total ejected particles Total received particles Percentage of reception Atmosphere 33 Al 1.39428×1016 9.04330008×1015 64.86% W 1.09506×1016 2.28100998×1015 20.83% Vacuum 33 Al 1.34463×1016 7.80557715×1015 58.05% W 8.71092×1015 1.965183552×1015 22.56% -
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