The work functions for metals and are 1.9eV and 5.0eV respectively. Perform necessary calculations…

Energy of the incident photon: $E = \frac{hc}{\lambda}$ Given $h = 6\cdot6 \times 10^{-34}\text{ J s}$, $c = 3 \times 10^{8}\text{ m s}^{-1}$, and $\lambda = 410\text{ nm} = 410 \times 10^{-9}\text{ m}$: $E = \frac{6\cdot6 \times 10^{-34}\text{ J s} \times 3 \times 10^{8}\text{ m s}^{-1}}{410 \times 10^{-9}\text{ m}} = 4\cdot83 \times 10^{-19}\text{ J}$ Converting to electron-volts ($1\text{ eV} = 1\cdot6 \times 10^{-19}\text{ J}$): $E = \frac{4\cdot83 \times 10^{-19}\text{ J}}{1\cdot6 \times 10^{-19}\text{ J eV}^{-1}} \approx 3\cdot02\text{ eV}$ Photoelectric emission occurs only when $E \ge W_{0}$: For metal $M_{1}$: $E = 3\cdot02\text{ eV} > W_{1} = 1\cdot9\text{ eV}$, hence metal $M_{1}$ will emit photoelectrons. For metal $M_{2}$: $E = 3\cdot02\text{ eV} < W_{2} = 5\cdot0\text{ eV}$, hence metal $M_{2}$ will not emit photoelectrons. Final answer: Metal M_1