V is plotted against 1/lambda where lambda is wavelength of incident radiations for two metals

Answer

A.

(i), (iii)

According to Einstein photoelectric equation,

       E = Kmax + ϕ

        h cλ = ϕ + eVs

∴     Vs =  hcλ - ϕ  1e

   Vs versus 1λ graph is a straight line.

    Slope 

           tanθ = h ce

  The curves intersect the 1λ axis, where Vs is zero

∴   h cλ - ϕ = 0

⇒   1λ = ϕh c

For metal 1

      ϕ1h c = 0.001

For metal 2

       ϕ2hc = 0.002

For metal 3

         ϕ3hc = 0.004

∴   ϕ1 : ϕ2 : ϕ3 = 1 : 2 : 4

For metal 1

    λO1 = hcϕ = 10.001 = 1000 nm

For metal 2, threshold wavelength,

     λO2 = hcϕ = 10.002 = 500 nm

Simillarly, 

      λO3 = hcϕ = 10.004 = 250 nm

          λvoilet = 400 nm

 For metal 1     

    λvoilet < λthreshold

  For metal 3 

           λvoilet >  λthreshold

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1. Metal 1 may be gold and metal 2 may be cesium
2. θ1 > θ2, if metal -1 is gold and metal -2 is cesium 
3. θ1 = θ2 for any two metals
4. θ1 > θ2, if metal -1 and metal -2 are gold and copper respectively