Question

The work functions (in eV) of Mg, Cu, Ag, Na respectively are 3.7, 4.8, 4.3, 2.3. From how many metals, the electrons will be ejected if their surfaces are irradiated with an electromagnetic radiation of wavelength 300 nm? (h = $6.6 \times 10^{-34}$ Js, 1 eV = $1.6 \times 10^{-19}$ J)

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Photoelectric effect: $E_{\text{photon}} = \frac{hc}{\lambda} \ge \text{Work function}$.

Answer 1

The Correct Option is C

For photoelectric emission to occur, the energy of the incident photon must be greater than or equal to the work function of the metal. The energy of a photon is given by $E = \frac{hc}{\lambda}$. $$ E = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{300 \times 10^{-9}} = 6.6 \times 10^{-19} \, \text{J} $$ Converting to eV: $$ E = \frac{6.6 \times 10^{-19}}{1.6 \times 10^{-19}} = 4.125 \, \text{eV} $$ The metals with work functions less than 4.125 eV are Mg (3.7 eV), Ag (4.3 eV - incorrect, greater than incident energy), Na (2.3 eV). Thus, electrons will be ejected from Mg and Na. Therefore, the answer is 2 (it appears the official key might be incorrect, perhaps assuming Ag's work function is lower than it actually is).