Answer:
1.26 × 10¹⁵ s⁻¹
Explanation:
Work function is the minimum energy required to remove an electron from the surface of metal
energy of the electron = hf - Φ
Φ = work function = hf₀ where f₀ = threshold frequency
f₀ = Φ / h where h ( Planck constant = 6.626 × 10⁻³⁴ Js)
Φ = 5.22eV = 5.22 × 1 eV where 1 eV = 1.60217662 × 10⁻¹⁹ J
Φ = 5.22 × 1.60217662 × 10⁻19 J = 8.363362 × 10⁻¹⁹ J
f₀ = (8.363362 ×10⁻¹⁹ J) / (6.626× 10⁻³⁴ Js) = 1.26 × 10¹⁵ s⁻¹
The frequency must be greater than the 1.26 × 10¹⁵ s⁻¹ to observe the emission
