The work function of a metal is 2.56 x 10-19J. Calculate the frequency of a photon whose energy is required to…

Analyze the question: This is a photoelectric effect problem. We need the photon frequency that ejects an electron with kinetic energy 3.0 eV from a metal with work function 2.56 × 10⁻¹⁹ J.

Concept: Einstein's photoelectric equation:

hf = φ + KE

where φ is the work function, KE is the maximum kinetic energy of the ejected electron, h is Planck's constant and f is the photon frequency.

Solution:

1. Convert the kinetic energy to joules: KE = 3.0 × 1.6 × 10⁻¹⁹ = 4.8 × 10⁻¹⁹ J

2. Add the work function: hf = 2.56 × 10⁻¹⁹ + 4.8 × 10⁻¹⁹ = 7.36 × 10⁻¹⁹ J

3. Solve for f: f = (7.36 × 10⁻¹⁹) / (6.6 × 10⁻³⁴) f ≈ 1.115 × 10¹⁵ Hz ≈ 1.12 × 10¹⁵ Hz

Why the others are wrong:

• D (8.42 × 10¹⁴ Hz) would come from using only a different energy combination, not the full φ + KE.
• A and B have powers of 10¹⁹, which are far too large for visible/UV photons.

Tip: In photoelectric questions, always convert eV to joules before using h in J·s. Then use hf = φ + KE. A quick check is that the frequency for UV light is around 10¹⁵ Hz, so an answer of 10¹⁹ Hz should look suspicious.