AIIMS 2018 Physics Threshold Frequency MCQ Question

Option B is correct because the energy of the photons (0.825 eV) must be compared to the work function of each metal, which is determined by their threshold frequencies. The work function (ϕ) can be calculated using the formula ϕ = hν₀, where h is Planck's constant (4.14 × 10⁻¹⁶ eV·s) and ν₀ is the threshold frequency. For metal A, the work function is approximately 0.72 eV (1.8 × 10¹⁴ Hz), which is less than 0.825 eV, allowing for photoemission. However, for metal B, the work function is approximately 0.91 eV (2.2 × 10¹⁴ Hz), which exceeds the photon energy, thus allowing only metal B to emit electrons. Option A is incorrect because although A can emit electrons, it is not the only metal that can do so, and option C is incorrect as A can emit electrons. Option D is incorrect because A and B cannot both emit electrons simultaneously given the energy constraints.