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本帖最后由 liaorongbao 于 2020-6-11 17:54 编辑
图片资料对应的DOI: 10.1007/0-306-47070-5_11
图片资料对应的题目: Multiphoton Excitation of Molecular Fluorophores and Nonlinear Laser Microscopy
作者:Chris Xu and Watt W. Webb
图片资料摘录:“For an electronic transition (wjf) in the visible frequency range and further assuming that the intermediate state and the final state are close in energy,we then have The one-photon absorption cross section of amolecule can be estimated by its dipole transition length (typically tofor a transition length of 10^-8 cm ). Hence, the estimated two-photonabsorption cross sections should be approximately 10^-49 * cm^4 * s/photon (Eq.11.1). ”
百度查可见光频率4.2*10^14到7.8*10^14。以下用τ表示希腊字母陶
问题1:体系中有三个能级,分别是Ei, Ej, Ef。intermediate state应该指对应Ej的这个虚设的能级。资料“ the intermediate state and the final state are close in energy”,意思是Ej和Ef近似吗?如果让Ej和Ef近似那么Ei和Ej不也是近似了吗?
问题2:The one-photon absorption cross section of a molecule can be estimated by its dipole transition length (typically σ ≈ 10^-16 to 10^-17 cm^2 for a transition length of 10^-8 cm )。这里能看出10^-8 cm数值的平方就是σ ≈ 10^-16 。 那么transition length(即为10^-8 cm)指什么?是不是吸光基团的长度?
问题3:图片文字描述双光子截面表达式中的τj=10^-15 to 10^-16 s。这个数值来源是直接取“可见光频率4.2*10^14到7.8*10^14”的倒数得到的吗?比如取6.0*10^14的倒数就是1.67*10^-15 s。
问题4:10^-49 * cm^4 * s/photon是怎么得到的?我的猜想:单光子截面的平方再乘以可见光频率的倒数。即为10^-17 cm^2 * 10^-17 cm^2 * 1.67*10^-15 s 就得到了10^-49 * cm^4 * s/photon。但是不巧的是尾巴上出现了量纲 cm^4 * s/photon而不是 cm^4 * s。为什么?
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