What is the minimum frequency of light required to observe the photoelectric effect on pd

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

c u c l 2 + 2 n a n o 3 → c u ( n o 3 ) 2 + 2 n a c l

  long weird answer but i hope this works

he would use the colour black