A light wave described by E = 60[sin(3 times 10^15t) + sin(12 times 10^15t)]$E = 60[\sin(3 \times 10^{15}t) + \sin(12 \times 10^{15}t)]$ (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately) ____ eV. (h = 6.6 times 10^-34text Jcdottexts$h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$. and e = 1.6 times 10^-19textC$e = 1.6 \times 10^{-19}\text{C}$)

### Related Formula K_textmax = hnu_textmax - phi_0$$K_{\text{max}} = h\nu_{\text{max}} - \phi_0$$ v = fracomega2pi$$v = \frac{\omega}{2\pi}$$ ### Core Logic The light wave consists of two frequencies governed by omega_1$\omega_1$ and omega_2$\omega_2$. omega_1 = 3 times 10^15text rad/s$\omega_1 = 3 \times 10^{15}\text{ rad/s}$ omega_2 = 12 times 10^15text rad/s$\omega_2 = 12 \times 10^{15}\text{ rad/s}$ The maximum kinetic energy of ejected photoelectrons will be determined by the highest frequency photon, which corresponds to omega_2 = 12 times 10^15text rad/s$\omega_2 = 12 \times 10^{15}\text{ rad/s}$. ### Step 1: Calculate Photon Energy Frequency nu_textmax = fracomega_22pi = frac12 times 10^152 times 3.14 approx 1.91 times 10^15text Hz$\nu_{\text{max}} = \frac{\omega_2}{2\pi} = \frac{12 \times 10^{15}}{2 \times 3.14} \approx 1.91 \times 10^{15}\text{ Hz}$. Energy of this photon: E_textphoton = hnu = (6.6 times 10^-34) times (1.91 times 10^15) = 1.26 times 10^-18text J$$E_{\text{photon}} = h\nu = (6.6 \times 10^{-34}) \times (1.91 \times 10^{15}) = 1.26 \times 10^{-18}\text{ J}$$ Convert this energy to eV: E_textmax = frac1.26 times 10^-181.6 times 10^-19 approx 7.87text eV approx 7.9text eV$$E_{\text{max}} = \frac{1.26 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 7.87\text{ eV} \approx 7.9\text{ eV}$$ ### Step 2: Calculate Maximum Kinetic Energy Using Einstein's photoelectric equation: K_textmax = E_textmax - phi_0$$K_{\text{max}} = E_{\text{max}} - \phi_0$$ K_textmax = 7.9 - 2.8 = 5.1text eV$$K_{\text{max}} = 7.9 - 2.8 = 5.1\text{ eV}$$ ### Pattern Recognition When a wave has multiple frequency components (E = E_1sinomega_1 t + E_2sinomega_2 t$E = E_1\sin\omega_1 t + E_2\sin\omega_2 t$), the K_textmax$K_{\text{max}}$ is always strictly determined by the highest frequency (highest energy) component. Ignore the lower frequency terms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Dual Nature of Radiation and Matter