Step 1: Understand the principles of the photoelectric effect

1. Condition for Photoemission: For the photoelectric effect to occur, the energy of an incident photon ($E$) must be greater than the work function ($\\\phi$) of the metal. The photon energy is given by $E = hf = \frac{hc}{\\\lambda}$.
2. Maximum Kinetic Energy: The maximum kinetic energy ($K_{max}$) of the ejected photoelectrons is given by Einstein's photoelectric equation: $K_{max} = E - \\\phi$.
3. Stopping Potential ($V_s$): The stopping potential is the negative potential required to stop the most energetic electrons. It is related to $K_{max}$ by $e V_s = K_{max}$. The magnitude of the stopping potential is determined by the highest energy photon that causes emission: $|V_s| = \frac{K_{max}}{e} = \frac{E_{max} - \\\phi}{e}$. If energies are expressed in eV, then $|V_s|$ in Volts is numerically equal to $K_{max}$ in eV.
4. Saturation Current ($I_s$): The saturation current is proportional to the number of photoelectrons emitted per second. This, in turn, is proportional to the number of incident photons per second with energy greater than the work function.

Step 2: Calculate the energy of incident photons

Given $hc = 1240~\mathrm{eV~nm}$, we calculate the energy for each wavelength:

• For $\\\\/lambda_1 = 550~\mathrm{nm}$: $E_1 = \frac{1240~\mathrm{eV~nm}}{550~\mathrm{nm}} \approx 2.25~\mathrm{eV}$
• For $\\\\/lambda_2 = 450~\mathrm{nm}$: $E_2 = \frac{1240~\mathrm{eV~nm}}{450~\mathrm{nm}} \approx 2.76~\mathrm{eV}$
• For $\\\\/lambda_3 = 350~\mathrm{nm}$: $E_3 = \frac{1240~\mathrm{eV~nm}}{350~\mathrm{nm}} \approx 3.54~\mathrm{eV}$

Step 3: Determine which wavelengths cause photoemission for each plate

We compare the photon energies with the work functions:

• Plate p ($\\\\/phi_p = 2.0~\mathrm{eV}$):

  • $E_1 = 2.25~\mathrm{eV} > \\\\phi_p$. Emission occurs.
  • $E_2 = 2.76~\mathrm{eV} > \\\\phi_p$. Emission occurs.
  • $E_3 = 3.54~\mathrm{eV} > \\\\phi_p$. Emission occurs.
  • All three wavelengths (550 nm, 450 nm, 350 nm) cause emission.

• Plate q ($\\\\/phi_q = 2.5~\mathrm{eV}$):

  • $E_1 = 2.25~\mathrm{eV} < \\\\phi_q$. No emission.
  • $E_2 = 2.76~\mathrm{eV} > \\\\phi_q$. Emission occurs.
  • $E_3 = 3.54~\mathrm{eV} > \\\\phi_q$. Emission occurs.
  • Two wavelengths (450 nm, 350 nm) cause emission.

• Plate r ($\\\\/phi_r = 3.0~\mathrm{eV}$):

  • $E_1 = 2.25~\mathrm{eV} < \\\\phi_r$. No emission.
  • $E_2 = 2.76~\mathrm{eV} < \\\\phi_r$. No emission.
  • $E_3 = 3.54~\mathrm{eV} > \\\\phi_r$. Emission occurs.
  • Only one wavelength (350 nm) causes emission.

Step 4: Compare the saturation currents ($I_s$)

The intensities of the three wavelengths are equal. Let this intensity be $I_0$. The number of photons per second ($N$) for a given wavelength $\\\\/lambda$ is given by $N = \frac{\text{Power}}{\text{Energy per photon}} = \frac{I_0 A}{hc/\\\\lambda} = \frac{I_0 A \\\\lambda}{hc}$. Thus, $N \propto \\\\lambda$. The saturation current is proportional to the sum of the number of photons (for wavelengths that cause emission).

• $I_{sp} \propto N_1 + N_2 + N_3 \propto 550 + 450 + 350 = 1350$ (relative units)
• $I_{sq} \propto N_2 + N_3 \propto 450 + 350 = 800$ (relative units)
• $I_{sr} \propto N_3 \propto 350$ (relative units)

From this, we conclude the order of saturation currents is $I_{sp} > I_{sq} > I_{sr}$.

Step 5: Compare the stopping potentials ($V_s$)

The stopping potential is determined by the highest energy photon that causes emission. In all three cases where emission occurs, the highest energy photon is from the 350 nm wavelength, i.e., $E_{max} = E_3 = 3.54~\mathrm{eV}$.

• For plate p: $|V_{sp}| = E_3 - \\\\phi_p = 3.54~\mathrm{eV} - 2.0~\mathrm{eV} = 1.54~\mathrm{V}$.
• For plate q: $|V_{sq}| = E_3 - \\\\phi_q = 3.54~\mathrm{eV} - 2.5~\mathrm{eV} = 1.04~\mathrm{V}$.
• For plate r: $|V_{sr}| = E_3 - \\\\phi_r = 3.54~\mathrm{eV} - 3.0~\mathrm{eV} = 0.54~\mathrm{V}$.

The order of the magnitudes of the stopping potentials is $|V_{sp}| > |V_{sq}| > |V_{sr}|$.

Step 6: Identify the correct I-V graph

We are looking for a graph with three curves representing p, q, and r that satisfy the following conditions:

1. Saturation Currents: The plateau for curve 'p' is the highest, 'q' is in the middle, and 'r' is the lowest ($I_{sp} > I_{sq} > I_{sr}$).
2. Stopping Potentials: The magnitude of the stopping potential (the x-intercept) for 'p' is the largest, 'q' is intermediate, and 'r' is the smallest ($|V_{sp}| > |V_{sq}| > |V_{sr}|$). This means the x-intercept for 'p' is the most negative.

The graph in option A correctly shows these relationships: the curve for p has the highest saturation current and the largest magnitude of stopping potential, the curve for q is intermediate in both, and the curve for r has the lowest saturation current and the smallest magnitude of stopping potential.