Light of energy 1.12 x 10 ^ -18 J is incident on a metal of ejected electrons with maximum energy of 8.0 x 10…

Analyze the question: This is a photoelectric effect problem. Light of energy E falls on a metal and the ejected electrons have a maximum kinetic energy. We need the threshold frequency f₀.

Concept: Einstein's photoelectric equation:

E = W₀ + KE_max, where W₀ = hf₀ (the work function)

So hf₀ = E − KE_max

Solution:

1. Work function: W₀ = 1.12 × 10⁻¹⁸ − 8.0 × 10⁻¹⁹ W₀ = 11.2 × 10⁻¹⁹ − 8.0 × 10⁻¹⁹ = 3.2 × 10⁻¹⁹ J

2. Threshold frequency: f₀ = W₀ / h = (3.2 × 10⁻¹⁹) / (6.626 × 10⁻³⁴) f₀ ≈ 0.483 × 10¹⁵ Hz ≈ 4.8 × 10¹⁴ Hz

Tip: Always write the energy values with the same power of ten (for example, 11.2 × 10⁻¹⁹ and 8.0 × 10⁻¹⁹) before subtracting. This avoids the most common slip in this type of question. Also remember: photon energy = work function + maximum kinetic energy.