Question

The work functions for potassium and caesium are 2.25 eV and 2.14 eV respectively. Will the photoelectric effect occur for either of these elements 

(i) with incident light of wavelength 5650 Å 

(ii) with light of wavelength 5180 Å?

(h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s, 

e = 1.6 × 10^-19 J, 

m (electron) = 9.1 × 10^-31 kg)

Answer 1

Data : Φ (potassium) = 2.25 eV

Φ (caesium) = 2.14 eV, 

λ₁ = 5650 Å = 5.650 × 10^-7 m, 

λ₂ = 5180 Å = 5.180 × 10^-7 m, 

h = 6.63 × 10^-34 J.s, 

c = 3 × 10⁸ m/s

Φ (potassium) = 2.25 eV

= 2.25 × 1.6 × 10^-19J =3.6 × 10^-19 J

Φ (caesium) = 2.14 eV = 2.14 × 1.6 × 10^-19 J 

= 3.424 × 10^-19 J

Photon energy, E = \(\cfrac{hc}λ\)

(i) For λ₁ = 5650 Å

This is greater than Φ (caesium), but less than Φ (potassium). Hence, photoelectric effect will occur in case of caesium, but not in case of potassium

(ii) For λ₂ = 5180 Å

This is greater than 0 for potassium and for caesium. Hence, photoelectric effect will occur in both the cases.