Minimum possible wavelength of emitted photoelectron is — Atomic Structure Chemistry Question

💡 Solution & Explanation

To minimize the de-Broglie wavelength of the emitted photoelectron (λ = h / √(2 × m × K.E.)), we need to maximize its kinetic energy. The energy of a single photon is E_p = Power / n_photons. The maximum possible photon energy is obtained by pairing the maximum power (5 W) with the minimum photon emission rate (4 × 10^18 s^-1): E_max = 5 / (4 × 10^18) = 1.25 × 10^-18 J. The maximum kinetic energy of the emitted photoelectron is K.E._max = E_max - Work Function = 1.25 × 10^-18 - 4.5 × 10^-19 = 8.0 × 10^-19 J. The minimum de-Broglie wavelength of the photoelectron is: λ_min = h / √(2 × m × K.E._max) = (6.626 × 10^-34) / √(2 × 9.11 × 10^-31 × 8.0 × 10^-19) ≈ 5.49 Å ≈ √30 Å. Thus, option (b) is correct.