The electric field at a point associated with a light wave is given by
E = 200 [sin (6 × 10¹⁵)t + sin (9 × 10¹⁵)t] Vm⁻¹
Given : h = 4.14 × 10⁻¹⁵ eVs
If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be

Solution

Physics sir · 3 steps

Two waves hit the metal, and the fastest electron comes from the higher frequency, so use only that one.

1. Pick the frequency. The light is a mix of two waves with angular frequencies 6 × 10¹⁵ rad s⁻¹ and 9 × 10¹⁵ rad s⁻¹. Each photon carries the energy of one of these frequencies. The maximum kinetic energy comes from the higher one, ω = 9 × 10¹⁵ rad s⁻¹.
2. Convert to frequency. The number inside the sine is angular frequency, so divide by 2π: ν = (9 × 10¹⁵)/2π = 1.432 × 10¹⁵ Hz.

   Kₘₐₓ = hν - φ = (4.14 × 10⁻¹⁵)(1.432 × 10¹⁵) - 2.50 = 5.93 - 2.50 eV

3. So. Kₘₐₓ ≈ 3.43 eV. The closest option, within rounding, is (d), 3.42 eV.

The idea to keep. In sin(ω t) the number is ω, not ν, so divide by 2π before using E = hν. When light has several frequencies, the highest one sets the maximum kinetic energy.

Answer from the NTA final answer key. Solution by PrepWiser.