Formulae for secondary electron yield from insulators and semiconductors
Ai-Gen Xie[1]*, Min Lai1, Yu-Lin Chen1, Yu-Qing Xia1
1School of Physics and Optoelectronic Engineering, Nanjing University of Information Science and Technology, Nanjing 210044, China
*Corresponding author E-mail:xagth@126.com
**This work was supported by the National Natural Science Foundation of China (No.11473049).
Abstract: The processes and characteristics of secondary electron emission in insulators and semiconductors were studied, and the formulae for the maximum yield (delta;m) at Wp0mle;800eV and the secondary electron yield from insulators and semiconductors delta; at the primary incident energy of 2 keVle;Wp0lt;10 keV (delta;2-10) and 10 keVle;Wp0le;100 keV (delta;10-100) were deduced. The calculation results were compared with their corresponding experimental data. It is concluded that the deduced formulae can be used to calculate delta;2-100 at Wp0mle;800eV.
Key words: Maximum secondary electron yield; Insulators and semiconductors; Secondary electron yield.
I. Introduction
Secondary electron yields from different materials have interested many authors [1-4]. The secondary electron yield from insulators and semiconductors (delta;) has found increasing applications in various areas, such as information technology, accelerator, scanning electron microscope, space flight etc. [5-7] However, due to the high resistance of insulators and semiconductors, it is difficult to overcome their surface charge problem and measure the secondary electron yield[8-10]. Thus, available delta; measurement data on insulators and semiconductors are not rich [11], especially in the incident energy range of 10–100 keV. Instead, many authors deduced formulae for key parameters to deduce delta; of insulators and semiconductors[12-16]; But up to now, no formulae is available for the incident energy range of 2–10 keV (delta;2-10) and 10–100 keV (delta;10-100), and for Wp0m, the incident energy at which delta; is maximized (delta;m).
In this study, based on secondary electron emission processes in insulators and semiconductors and the formulae of delta;2-10, delta;10-100, and the maximum yield delta;m at Wp0mle;800 eV were deduced, involving the primary range of 10 keVle;Wp0le;100 keV (R10-100) and 2 keVle;Wp0lt;10 keV (R2-10), the atomic number Z, and the high energy back-scattering coefficient r, apart from the different forms of secondary electron yield . For compounds, Z represents average atomic number of compounds, for example, Z = (20 8)/2 for CaO; and r represents back-scattering coefficient in the energy range of Wp0ge;10 keV.
II. Methods
2.1 Formula for delta;m
When electrons enter an insulator or semiconductor, secondary electrons are generated due to energy deposition of the primary electrons. Suppose that N(x,Wp) is the number of secondary electrons produced at a depth x and primary electron energy of Wp, N(x,Wp) is proportional to average loss of primary electrons per unit path length [17,18]:
, (1)
where Wp is primary energy at given depth in insulator or semiconductor, S is the path length of primary electrons, and ε is the average energy required to produce an internal secondary electron in an insulator or semiconductor.
The probability of an internal secondary electron reaching the surface of the insulator or semiconductor and passing over the surface barrier into vacuum can be written as [17,18]:
f(x)= Bexp(minus;alpha;x) , (2)
where alpha; is the absorption coefficient (and 1/alpha; is mean escape depth of secondary electrons), and B is the probability that an internal secondary electron escapes into vacuum upon reaching the surface of insulator or semiconductor.
From Eqs. (1) and (2), the yield due to primary electron can be written as[19]:
, (3)
According to Seiler [17], the maximum yield due to primary electron delta;pm is at R = 2.3/alpha;. Then, delta;pm can be written as [19]:
. (4)
At Wp0mle;800 eV, the primary range R can be expressed as [20]
R = 2times;10minus;9Aalpha;Wp0m/(rho;Z2/3) , (5)
where rho; is material density, Z is atomic number and Aalpha; is atomic weight.
It can be assumed that the R at Wp0mle;800 eV approximately equals t
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绝缘体和半导体的二次电子发射系数公式
谢爱根,赖敏,陈玉林,夏雨晴
物理与光电工程学院, 南京信息工程大学, 南京 210044,中国.
摘要:研究了绝缘子和半导体二次电子发射的过程和特性,得到了Wp0mle;800eV时的最大系数(delta;m)和2 keV初次入射能量下绝缘子和半导体的二次电子系数delta; 推导出Wp0 lt;10keV(delta;2-10)和10keVle;Wp0le;100keV(delta;10-100)。 计算结果与其相应的实验数据进行比较。 得出结论:推导的公式可以用来计算Wp0mle;800eV的delta;2-100。
关键词:最大二次电子系数; 绝缘子和半导体; 二次电子系数。
I. 介绍
来自不同材料的二次电子产量已经引起许多作者的兴趣[1-4]。绝缘体和半导体的二次电子系数(delta;)在信息技术,加速器,扫描电子显微镜,空间飞行等各个领域得到了越来越多的应用[5-7]。然而,由于绝缘体和半导体的高电阻,很难克服它们的表面电荷问题并测量二次电子系数[8-10]。因此,绝缘体和半导体的可用delta;测量数据并不丰富[11],特别是在10-100 keV的入射能量范围内。相反,许多作者推导出了推导绝缘体和半导体delta;的关键参数的公式[12-16];但到目前为止,对于2-10keV(delta;2-10)和10-100keV(delta;10-100)的入射能量范围没有公式可用,对于Wp0m,delta;最大化的入射能量(delta;m) 。
在本研究中,基于绝缘子和半导体中的二次电子发射过程,推导出Wp0mle;800eV的delta;2-10,delta;10-100和最大系数delta;m的公式,其中主要范围为10keVle;Wp0le;100除了不同形式的二次电子系数之外,keV(R10-100)和2keVle;Wp0lt;10keV(R2-10),原子序数Z和高能量后向散射系数r。对于化合物,Z表示化合物的平均原子数,例如,对于CaO,Z =(20 8)/ 2; r表示Wp0ge;10keV能量范围内的反向散射系数。
II. 方法
2.1 公式delta;m
当电子进入绝缘体或半导体时,由于一次电子的能量沉积而产生二次电子。 假设N(x,Wp)是在深度x和Wp的一次电子能量产生的二次电子的数量,N(x,Wp)与每单位路径长度的一次电子的平均损失成比例[17,18]:
, (1)
其中Wp是绝缘体或半导体中给定深度处的一次能量,S是一次电子的路径长度,并且ε是在绝缘体或半导体中产生内部二次电子所需的平均能量。
内部二次电子到达绝缘体或半导体表面并通过表面障壁进入真空的可能性可以写成:[17,18]:
f(x)= Bexp(minus;alpha;x) , (2)
其中alpha;是吸收系数(1 /alpha;是二次电子的平均逸出深度),B是内部二次电子在到达绝缘体或半导体表面时逸入真空的概率。
从方程 (1)和(2),由一次电子产生的收益可写为[19]:
, (3)
根据塞勒[17],由于一次电子的最大收益delta;pm在R = 2.3 /alpha;。 然后,delta;pm可以写成[19]:
. (4)
在Wp0mle;800eV时,主要范围R可以表示为[20]
R = 2times;10minus;9Aalpha;Wp0m/(rho;Z2/3) , (5)
其中rho;是材料密度,Z是原子序数,Aalpha;是原子量。
可以假设Wp0mle;800eV处的R近似等于对应的S,推导出delta;m的公式。 因此,Wp0mle;800eV时一次电子的单位路径长度的能量损失可以通过对Eq。(5)
dWp/dS = minus;rho;Z2/3/(2times;10minus;9Aalpha;) . (6)
Wp0mle;800eV时的delta;pm可以通过结合方程式获得。 (4) and (6)
delta;pm = (e2.3minus;1)Brho;Z2/3/(2times;10minus;9Aalpha;alpha;e2.3ε) . (7)
可以写出delta;m和delta;pm之间的关系as [21]:
delta;m = (1 1.26r)delta;pm . (8)
对于给定的材料,r是一个常数[22]。 delta;m可以通过组合等式来获得。(7) and (8).
delta;m = (1 1.26r)(e2.3minus;1)Brho;Z2/3/(2times;10minus;9Aalpha;alpha;e2.3ε) (9)
2.2 公式delta;10-100
当10keVle;Wp0le;100keV的一次电子进入绝缘体或半导体时,R10-100可以表示为[20]
. (10)
可以假设R10-100近似等于相应的S来推导delta;10-100的公式。 因此,通过区分方程式可以得到下面的表达式 (10)
. (11)
R10-100比二次电子T的最大逸出深度大得多。例如,在Wp0中Si(Z = 14,rho;= 2.35g / cm3,Aalpha;= 28.1)[13] = 10 keV,我们有R10 =16055Aring;和T =41-54Aring;[23]。 因此,大部分初级能量在T外耗散,并且T内部的一次能量变化很小。然后,在T内每单位路径长度的10keVle;Wp0le;100keV能量范围内一次电子的能量损失可以近似为 写为 .
(12)
在10keVle;Wp0le;100keV的能量范围内的delta;p可以通过结合方程式来获得。 (3)和(12) .
(13)
R10-100比T大得多。在T外部激发的内部二次电子不能被发射到真空中[23],并且T近似等于5 /alpha;[23]。 因此,方程(13)的定积分[0,R]可以用[0,5 /alpha;]代替。 然后,我们有:
. (14)
delta;由delta;p和由于背散射电子delta;r产生的产量组成,即[24]
. (15)
其中beta;是一个背散射电子的平均二次电子产生与一个一次电子的平均二次电子产生的比率,并且eta;是Wp0处的后向散射系数。 由于背散射电子的平均发射角对于二次电子的激发比一次电子的正常入射更有利,并且背散射电子的平均能量小于W p0,所以beta;大于1 [17]。
根据理论和实验结果,对于10keVle;Wp0le;100keV,金属beta;约为2(beta;10-100metalasymp;2)[25-28].
因为来自绝缘体或半导体的背散射电子的平均发射角和平均能量类似于来自金属的背散射电子[28-29],
我们假设beta;10-100绝缘体asymp;2对于10keVle;Wp0le;100keV。 因此,我们有:
delta;10–100= (1 2r)delta;p . (16)
能量范围10keVle;Wp0le;100keV和Wp0mle;800eV的delta;可以通过结合方程式得到。 (9),(14)和(16).
(17)
2.3 公式delta;2-10
当2keVle;Wp0lt;10keV的一次电子进入绝缘体或半导体时,R2-10可以表示为[20]
. (18)
可以假设在推导delta;2-10的公式期间R2-10近似等于相应的S。 因此,我们有方程 (19)
. (19)
在Wp0m lt;800 eV的能量范围内,R2-10比T大得多。 例如,对于Si(Z = 14,rho;= 2.35g / cm3,Aalpha;= 28.1)[13],在Wp0 = 2000eV时,用方程 (18)我们有R2 = 1222?和T = 41-54?[23]。 因此,大部分一次能量消散在T外,一次能量在T内部只有很小的变化。那么,在T能量范围内,2keV能量范围内一次电子的能量损失le;Wp0lt;10keV /单位路径长度 大致写为: . (20)
根据方程式 (3)和(20)中,能量范围2keVle;Wp0lt;10 keV的delta;p可写为
. (21)
在Wp0m lt;800 eV的能量范围内,R2-10远大于T,在T外激发的内部二次电子不能发射到真空[23],而Tasymp;5/alpha;[23]
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