Probing Local Strain at MX2−Metal Boundaries with Surface Plasmon-Enhanced Raman Scattering

Probing Local Strain at MX2−Metal Boundaries with Surface Plasmon-Enhanced Raman Scattering 1 1. Optical images, Raman spectra, and PL spectra of exfoliated monolayer MoS2 without and with 5 nm Ag deposition. monolayer monolayer Figure S1. (a) Optical image of pristine exfoliated MoS2 monolayer and few layers. (b) Corresponding optical image of MoS2 layers after 5 nm Ag deposition. (c) Raman spectra of pristine MoS2 monolayer, Ag-coated monolayer, Ag-coated bulk, and Ag-coated SiO2/Si substrate. The nominal Ag thickness is 5 nm. The E monolayer. The energy splitting of E mode splits into two peaks for Ag-coated Raman mode can be estimated to be 7.5 cm-1 from a Lorentzian multi-peak fitting. The spectra are shifted vertically for clarity. (d) PL spectra of pristine monolayer and Ag-coated monolayer. A shoulder PL peak appears at an energy of ~0.15 eV below the A exciton PL. 2 2. Raman and PL spectra of monolayer and bilayer MoS2 with 1 nm Au deposition. Figure S2. (a) Raman spectra of Au-coated monolayer and bilayer MoS2 excited by a 488 nm laser. The nominal Au thickness is 1 nm. It shows that the E Raman mode also splits for Au- coated monolayer, and the energy splitting is estimated to be 4.4 cm-1 from a Lorentzian multipeak fitting. (b) PL spectra of Au-coated monolayer and bilayer MoS2. 3. Reflection spectrum of Ag NPs and Raman spectra of Ag/MoS2 at different excitation wavelengths. 3 Figure S3. (a) Reflection spectrum of Ag nanoparticles on monolayer MoS2. (b) Raman spectra of Ag-coated monolayer MoS2 at different excitation wavelengths. The thickness of Ag is 1 nm nominally. 4. Analysis of strain distribution in monolayer and bilayer MoS2 coated with Ag NP. 4 Figure S4. Force analysis and diagram of strain relaxation in monolayer (a) and bilayer (b) MoS2 deposited with Ag NP. In Figure S4, we provide a qualitative model for the strain distribution in monolayer and bilayer MoS2. Local force analysis for the element nearest to the Ag NP and the proximate elements are shown in Fig. S4a for monolayer. F0 is the force coming from the intrinsic stress. Fi ( ) is the internal force in monolayer, which monotonously decreases until the strain is fully relaxed. Fsub ( ) is the external force from substrate, representing the load transfer to substrate. Similarly, the local force analysis for the elements in both top layer and bottom layer is shown in Fig. S4b for bilayer. Fv ( ) is the interlayer friction coming from interlayer van de Waals force, representing the load transfer from the top to the bottom layer. F2 ( ) is the internal force in bottom layer, which also monotonously decreases until the strain is fully relaxed. For the monolayer MoS2 or the top layer of bilayer MoS2, the transfer equation of intrinsic stress can be described as / =( − )/(∆ ℎ) = ( )/(∆ ℎ) =  /ℎ ∙ where A is the cross-section area, ℎ is the thickness of monolayer, ∆ (Eq. S1) is the width of the selected element,  is the friction coefficient between the top layer and the underneath interface, is the van der Walls force in an unit area of the top layer, and dx is the slide distance at that position (the last step is only true up to the sliding point). Equation S1 indicates that the propagation of intrinsic stress with distance strongly depends on  /ℎ. According to the two dimensional Young’s modulus measurement,2 the modulus of 5 bilayer MoS2 is much lower than twice the value of monolayer, probably due to the interlayer sliding, which suggests that the interaction between top and bottom layers of bilayer is weaker than the interaction from substrate. In monolayer MoS2, therefore,  are larger than the value in the top layer of bilayer MoS2 because of the strong clamping by the substrate. As a result, the strain relaxes faster with distance in monolayer MoS2 than in the top layer of bilayer MoS2. In the bottom layer of bilayer MoS2, stress transfer can be similarly described as / = ( − )/ℎ ∙ (Eq. S2) However, because the initial local stress in the bottom layer is much lower than the top layer due to the weak van der Waals interaction, the strain in the bottom layer should be much smaller than that in the top layer. Curves of strain distribution are also illustrated in Fig. S4. 6