The work function () of some metals is listed below. The number of metals which will show photoelect — Atomic Structure Chemistry Question

💡 Solution & Explanation

Energy of incident photon $E = \frac{1240 \text{ eV}\cdot\text{nm}}{300 \text{ nm}} \approx 4.133$ eV. For photoelectric emission to occur, the incident energy must be greater than the work function ($E > \phi$). The metals satisfying this condition are Li (2.4 eV), Na (2.3 eV), K (2.2 eV), and Mg (3.7 eV). There are 4 such metals. Therefore, correct answer is 4.