高频单线圈双调谐NMR单晶探头的研制

本文报道了一种测角单晶NMR探头。该探头采用单线圃双调谐电路,工作频率在90MHz-110 MHz连续可调,可进行交叉极化大功率去耦实验。文中提出了一种简单而有效的测角装置,可使单晶绕三个互相垂直轴转动实现单晶的NMR测量。作为典型的应用例子,本文利用该探头实现了单晶DGO(2NH₂CH₂COOH·H₂C₂O₄)屏蔽张量的测量。     

Field Emission from Silicon Nanocrystallite Films with Compact Alignment and Uniform Orientation

Patterned silicon nanocrystallite (SINC) films were fabricated on (100) orientation p-type boron-doped sificon wafer by the hydrogen ion implantation technique and the anodic etching method. The efficient field emission with low turn-on field of about 3.5V/μm at current density of 0.1μA/cm^2 was obtained. The emission current density from the SiNC films reached 1mA/cm^2 under a bias field of about 9.1V/μm. The experimental results demonstrate that there are great potential applications of the SiNC films for fiat panel displays. A surface treatment with hydrogen plasma was performed on the SiNC films and a significant improvement of emission properties was achieved.     

Effect of Incident Intensity on Films Growth in Pulsed Laser Deposition

Incident intensity, defined by the amount of particles deposited per pulse, is an important parameter in the film growth process of pulsed laser deposition （PLD）. Different from previous models, we investigate the irreversible and reversible growth processes by using a kinetic Monte Carlo method and find that island density and film morphology strongly depend on pulse intensity. At higher pulse intensities, lots of adatoms instantaneously diffuse on the substrate surface, and then nucleation easily occurs between the moving adatoms resulting in more smaller-size islands. In contrast, at the lower pulse intensities, nucleation event occurs preferentially between the single adatom and existing islands rather than forming new islands, and therefore the average island size becomes larger in this case. Additionally, our results show that substrate temperature plays an important role in film growth. In particular, it can determine the films shape and weaken the effect of pulse intensity on film growth at the lower temperatures by controlling the mobility rate of atoms. Our results can match the related theoretical and experimental results.     

关于新旧教材比较及对教学的启迪--以万有引力定律教学为案例

l比较新旧教材的意义 对教师而言，比较新旧教材，无疑是教学活动中不可缺少的环节．通过新旧教材的对比，可以使教师掌握教材编排．更重要的是，通过新旧教材对比，让教师更好地领会新旧教材的不同，进而了解编者的意图，体会新课程标准的理念，从而在教学活动中更好地理解和落实新课程标准．下面，笔者以2004年初审通过的广东教育出版社出版的《普通高中课程标准实验教科书物理（必修2）》第三章第一节“万有引力定律”，     

一类具有二维分叉方程的积分方程的分叉

本文研究一类含参数的非线性积分方程的分叉问题,其中的积分算子的线性化算子在分叉值点处有二维零空间。利用Liapunov-Schmidt约化方法和基于系统的对称性的群论方法,得到了周期分叉解存在的充分条件。     

Numerical Study on Left-Handed Materials Made of Ferrite and Metallic Wires

Due to coupling effect, we show that it is difficult to realize the left-handed material by placing metallic wires directly into a ferrite matrix. However by introducing an insulating material round the metallic wires to decouple the direct interaction between the metallic wire and ferrite matrix, we have proposed two microstructures, which are shown by numerical simulation to have negative refractive indexes. The influence of microstructure on the transmission property is also examined.     

A Comparison of Different Measures for Dynamical Event Mean Transverse Momentum Fluctuation

Various measures for the dynamical event mean transverse momentum fluctuation are compared with the real dynamical fluctuation using a Monte Carlo model. The variance calculated from the G-moments can reproduce the dynamical variance well, while those obtained by subtraction procedures are approximate measures for not very low multiplicity. Фpt, proposed by Gadzicki M and Mrowczynski S [Z. Phys. C 54 (1992)127], can also serve as an approximate measure after being divided by the square root of mean multiplicity.     

一个数学问题的再探讨

命题若复数z_1,z_2,z_3满足z_1+z_2+z_3=0,|z_1|=|z_2|=|z_3|=1,则复平面内以z_1,z_2,z_3所对应的点为顶点的三角形是内接于单位圆的正三角形。文[1]的作者给出了该命题的一种证法。并探讨了该命题的逆命题。若复平面内以模为1的复数z_1,z_2,z_3所对应的点为顶点的三角形是正三角形,则z_1+z_2+z_3=0。容易证明此命题也正确(略)。作者还对该命题进行了推广,笔者读后受益非浅。本文将进一步探讨以上两个命题在解题中的应用。下面以例示明。例1 (1986年苏州市数学竞赛题) 已知复数z满足|z|=1,z~(11)+z=1,求z。解∵ |z|=1, ∴|z~(11)|=|z|=|-1|=1 又z~(11)+z+(-1)=0 ∴z~(11),z,-1所对应的三点构成一个正三角形。故z=(-1)(cos120°±sin120°)=(1/2)±3~(1/2)/2i 例2 (1987年第二届全国高中数学冬令营赛题)