1. **Problem statement:** Calculate the threshold frequency $f_0$ of a metal with work function $\phi = 2.6$ eV. Calculate the maximum speed $v_{max}$ of electrons emitted when light of wavelength $\lambda = 440$ nm hits the metal. 2. **Relevant formulas and concepts:** - Work function $\phi$ relates to threshold frequency by $\phi = hf_0$, where $h = 6.626 \times 10^{-34}$ Js is Planck's constant. - Photon energy $E = hf = \frac{hc}{\lambda}$, where $c = 3.0 \times 10^8$ m/s is speed of light. - Maximum kinetic energy of emitted electrons: $K_{max} = hf - \phi$. - Kinetic energy relates to speed by $K_{max} = \frac{1}{2}mv_{max}^2$, where $m = 9.11 \times 10^{-31}$ kg is electron mass. - 1 eV = $1.602 \times 10^{-19}$ J. 3. **Calculate threshold frequency $f_0$:** Convert work function to joules: $$\phi = 2.6 \text{ eV} = 2.6 \times 1.602 \times 10^{-19} = 4.1652 \times 10^{-19} \text{ J}$$ Use $\phi = hf_0$ to find $f_0$: $$f_0 = \frac{\phi}{h} = \frac{4.1652 \times 10^{-19}}{6.626 \times 10^{-34}} = 6.28 \times 10^{14} \text{ Hz}$$ 4. **Calculate photon energy $E$ for $\lambda = 440$ nm:** Convert wavelength to meters: $$\lambda = 440 \text{ nm} = 440 \times 10^{-9} = 4.4 \times 10^{-7} \text{ m}$$ Calculate photon energy: $$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{4.4 \times 10^{-7}} = 4.52 \times 10^{-19} \text{ J}$$ 5. **Calculate maximum kinetic energy $K_{max}$:** $$K_{max} = E - \phi = 4.52 \times 10^{-19} - 4.1652 \times 10^{-19} = 3.55 \times 10^{-20} \text{ J}$$ 6. **Calculate maximum speed $v_{max}$:** Use $K_{max} = \frac{1}{2}mv_{max}^2$: $$v_{max} = \sqrt{\frac{2K_{max}}{m}} = \sqrt{\frac{2 \times 3.55 \times 10^{-20}}{9.11 \times 10^{-31}}} = \sqrt{7.8 \times 10^{10}} = 8.83 \times 10^{5} \text{ m/s}$$ **Final answers:** - Threshold frequency $f_0 = 6.28 \times 10^{14}$ Hz - Maximum speed of emitted electrons $v_{max} = 8.83 \times 10^{5}$ m/s