A.

(i), (iii)

According to Einstein photoelectric equation,

       E = K_max + $\mathrm{\varphi }$

        $\frac{\mathrm{h} \mathrm{c}}{\mathrm{\lambda }}$ = $\mathrm{\varphi }$+ eV_s

∴     V_s = $\left[ \frac{\mathrm{hc}}{\mathrm{\lambda }} - \mathrm{\varphi } \right]$ $\frac{1}{\mathrm{e}}$

   V_s versus $\frac{1}{\mathrm{\lambda }}$ graph is a straight line.

    Slope 

           $\mathrm{tan\theta } = \frac{\mathrm{h} \mathrm{c}}{\mathrm{e}}$

  The curves intersect the $\frac{1}{\mathrm{\lambda }}$ axis, where V_s is zero

∴   $\frac{\mathrm{h} \mathrm{c}}{\mathrm{\lambda }} - \mathrm{\varphi }$ = 0

⇒   $\frac{1}{\mathrm{\lambda }}$ = $\frac{\mathrm{\varphi }}{\mathrm{h} \mathrm{c}}$

For metal 1

      $\frac{{\mathrm{\varphi }}_1}{\mathrm{h} \mathrm{c}} = 0.001$

For metal 2

       $\frac{{\mathrm{\varphi }}_2}{\mathrm{hc}}$ = 0.002

For metal 3

         $\frac{{\mathrm{\varphi }}_3}{\mathrm{hc}}$ = 0.004

∴   ${\mathrm{\varphi }}_1 : {\mathrm{\varphi }}_2 : {\mathrm{\varphi }}_3$ = 1 : 2 : 4

For metal 1

    ${\mathrm{\lambda }}_{{\mathrm{O}}_1} = \frac{\mathrm{hc}}{\mathrm{\varphi }} = \frac{1}{0.001}$ = 1000 nm

For metal 2, threshold wavelength,

     ${\mathrm{\lambda }}_{{\mathrm{O}}_2} = \frac{\mathrm{hc}}{\mathrm{\varphi }}$ = $\frac{1}{0.002}$ = 500 nm

Simillarly, 

      ${\mathrm{\lambda }}_{{\mathrm{O}}_3} = \frac{\mathrm{hc}}{\mathrm{\varphi }} = \frac{1}{0.004}$ = 250 nm

          λ_voilet = 400 nm

 For metal 1     

    λ_voilet < λ_threshold

  For metal 3 

           λ_voilet >  λ_threshold

C.

10.2 eV

Excitation energy is defined as the energy required to take the electron from ground level orbit to any higher order orbit (ie, n = 2,3,4...)

Given, ground state energy of hydrogen atom

E₁ = - 13.6/(2)² eV

Energy of electron in first excited state (ie, n = 2)
E₂ = - 13.6/ (2)² eV

Therefore, excitation energy

ΔE = E₂-E₁
 =(-13.6/4) - (-13.6)
 = - 3.4 + 13.6 = 10.2 eV

A.

1.6 × 10^-17 J

The energy transferred to electrons is e V joules. So each electron gains kinetic energy equal to the amount of energy transferred from the electrical supply. The electron starts from rest ( near enough ) so the Kinetic energy gained is given by 1 / 2 mv² where m is the mass of electron and v is speed.

KE = e V

    = 1.6 × 10^-19 × 100

KE = 1.6 × 10^-17 J

C.

0.6 eV

The stopping potential depends on the kinetic energy of the electrons, which is affected only by the frequency of incoming light and not by its intensity.

The maximum kinetic energy of the photoelectron
(KE)_max = E - Φ  =1.8-1.2
            =  0.6  eV
i.e   eV_o = (KE)_max

or    eV_o = 0.6 eV

∴ Stopping potential V_o = 0.6 V

A.

30 kV

From the formula

$$\begin{gathered} \mathrm{e} \mathrm{V} =\mathrm{h}\frac{\mathrm{c}}{\mathrm{\lambda }} \\ \mathrm{V} =\frac{\mathrm{h} \mathrm{c}}{\mathrm{e} \mathrm{\lambda }} \\ =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 0.4125\times {10}^{-10}} \\ =30 \times {10}^3 \mathrm{volt} \\ =30 \mathrm{kvolt} \end{gathered}$$