### Related Formula Einstein's Photoelectric Equation: K_textmax = hnu - phi = frachclambda - phi$K_{\text{max}} = h\nu - \phi = \frac{hc}{\lambda} - \phi$ For emission to occur, the photon energy must exceed the work function: E_textphoton > phi implies lambda < lambda_textthreshold = frachcphi$E_{\text{photon}} > \phi \implies \lambda < \lambda_{\text{threshold}} = \frac{hc}{\phi}$ Useful shortcut: hc approx 1240mathrm~eVcdot nm$hc \approx 1240\mathrm{~eV\cdot nm}$. ### Core Logic Let's find the threshold wavelength lambda_textthreshold$\lambda_{\text{threshold}}$ for a work function of phi = 3mathrm~eV$\phi = 3\mathrm{~eV}$: lambda_textthreshold = frac1240mathrm~eVcdot nm3mathrm~eV approx 413.3mathrm~nm$\lambda_{\text{threshold}} = \frac{1240\mathrm{~eV\cdot nm}}{3\mathrm{~eV}} \approx 413.3\mathrm{~nm}$ To cause photoelectric emission, the wavelength of the incident light must be shorter than this threshold: lambda < 413.3mathrm~nm$\lambda < 413.3\mathrm{~nm}$ ### Step 1: Comparing Visible Spectrum Colors Let's look at the standard approximate wavelength ranges for visible light: - **Red**: 620 - 750mathrm~nm$620 - 750\mathrm{~nm}$ (Energy approx 1.65 - 2.0mathrm~eV$\approx 1.65 - 2.0\mathrm{~eV}$) - **Yellow**: 570 - 590mathrm~nm$570 - 590\mathrm{~nm}$ (Energy approx 2.1 - 2.2mathrm~eV$\approx 2.1 - 2.2\mathrm{~eV}$) - **Green**: 495 - 570mathrm~nm$495 - 570\mathrm{~nm}$ (Energy approx 2.2 - 2.5mathrm~eV$\approx 2.2 - 2.5\mathrm{~eV}$) - **Blue**: 450 - 495mathrm~nm$450 - 495\mathrm{~nm}$ (At lower end, approaching violet down to 380mathrm~nm$380\mathrm{~nm}$; Energy approx 2.5 - 3.3mathrm~eV$\approx 2.5 - 3.3\mathrm{~eV}$) Only **Blue** light contains wavelengths extending below 413.3mathrm~nm$413.3\mathrm{~nm}$ (high enough photon energy to surpass 3mathrm~eV$3\mathrm{~eV}$). Therefore, blue light is required to cause photoelectric emission from this metal. ### Pattern Recognition Remember standard photon energies of visible colors: Blue/Violet photons have higher energy (typically > 2.8mathrm~eV$> 2.8\mathrm{~eV}$), while Red/Yellow photons have much lower energy (< 2.2mathrm~eV$< 2.2\mathrm{~eV}$). For a higher work function like 3mathrm~eV$3\mathrm{~eV}$, only highly energetic blue/violet light can succeed. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter