When a surface 1cm thick is illuminated with light of wave length λ the stopping potential is Vo, but when…..

Q. When a surface 1cm thick is illuminated with light of wave length λ the stopping potential is V₀, but when the same surface is illuminated by light of wavelength 3λ, the stopping potential is V₀ /6 .The threshold wavelength for metallic surface is:

(a) 4λ

(b) 5λ

(c) 3λ

(d) 2λ

Ans: b

Sol: hc/λ = φ + eV₀      …..(i)

hc/3λ = φ + eV₀ /6       …..(ii)

On dividing ,(i)/(ii)

3 = φ + eV₀ /(φ + eV₀ /6 )

3φ + eV₀ /2 = φ + eV₀

2φ = eV₀ /2 ⇒ φ = eV₀ /4

⇒ hc/λ₀ = eV₀ /4   ⇒ eV₀ = 4hc/λ₀ —–(iii)

Subtracting , (i)−(ii)

hc/λ(1−1/3) = eV₀(1−1/6)

(hc/λ)×2/3 = eV₀ ×5/6

eV₀  = 4hc/5λ   ……(iv)

4hc/λ₀ = 4hc/5λ  from (iii)

λ₀ = 5λ