Question

When a radiation of wavelength 3000 Å is incident on metal, 1.85 V stopping potential is obtained. What will be threshed wave length of metal? {h = 66 × 10^–34 J.s, c = 3 × 10⁸ (m/s)} 

(A) 4539 Å 

(B) 3954 Å 

(C) 5439 Å 

(D) 4395 Å

Answer 1

The correct option is (C) 5439 Å.

Explanation:

Energy = eV = (hc / λ) – ɸ. hence ɸ = (hc / λ) – eV

 ɸ = [{6.6 × 10^–34 × 3 × 10⁸} / {3000 × 10^–10}] – (1.85 × 1.6 × 10^–19)

= 6.6 × 10^–19 – 2.96 × 10^–19

= 3.64 × 10^–19 J

ɸ  = 2.275 eV ----- 1eV = 1.6 × 10^–19 J

Now ɸ = (hc / λ_o) hence λ_o = (hc / ɸ)

∴ λ_o = {(6.6 × 10^–34 × 3 × 10⁸) / (2.275 × 1.6 × 10^–19)}

= 5.439 × 10^–7 m

λ_o = 5439 Å