Photoelectric effect experiments are performed using three different metal plates p,q and r having work functions `phi_p=2.0eV,and `phi_q=2.5eV `and `phi_(tau) `= 3eV` and wavelengths of 550 nm,550nm, and 350 nm with equal intensities , illuminates each of the plates . The correct I - V graph for the experiment is (Take hc = 1240 eV nm)

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Explanation

To solve the problem regarding the photoelectric effect experiments with three different metal plates (P, Q, and R), we need to analyze the given data and apply the principles of the photoelectric effect. Here's a step-by-step solution:

### Step 1: Understand the Work Functions and Wavelengths
We have three metal plates with the following work functions:
- Plate P: \( \phi_p = 2.0 \, \text{eV} \)
- Plate Q: \( \phi_q = 2.5 \, \text{eV} \)
- Plate R: \( \phi_r = 3.0 \, \text{eV} \)

The wavelengths of light illuminating these plates are:
- Plate P: \( \lambda_p = 550 \, \text{nm} \)
- Plate Q: \( \lambda_q = 550 \, \text{nm} \)
- Plate R: \( \lambda_r = 350 \, \text{nm} \)

### Step 2: Calculate the Energy of the Incident Photons
Using the formula for the energy of a photon:
\[ E = \frac{hc}{\lambda} \]
where \( hc = 1240 \, \text{eV} \cdot \text{nm} \).

**For Plate P and Q (550 nm):**
\[ E_p = E_q = \frac{1240}{550} \approx 2.25 \, \text{eV} \]

**For Plate R (350 nm):**
\[ E_r = \frac{1240}{350} \approx 3.54 \, \text{eV} \]

### Step 3: Calculate the Maximum Kinetic Energy
The maximum kinetic energy of the emitted electrons can be calculated using:
\[ KE_{max} = E - \phi \]

**For Plate P:**
\[ KE_{max, P} = E_p - \phi_p = 2.25 - 2.0 = 0.25 \, \text{eV} \]

**For Plate Q:**
\[ KE_{max, Q} = E_q - \phi_q = 2.25 - 2.5 = -0.25 \, \text{eV} \quad (\text{no photoelectrons emitted}) \]

**For Plate R:**
\[ KE_{max, R} = E_r - \phi_r = 3.54 - 3.0 = 0.54 \, \text{eV} \]

### Step 4: Calculate the Stopping Potential
The stopping potential \( V_0 \) can be calculated using:
\[ V_0 = \frac{KE_{max}}{e} \]

**For Plate P:**
\[ V_{0,P} = 0.25 \, \text{V} \]

**For Plate Q:**
\[ V_{0,Q} = \text{no emission} \quad (V_{0,Q} = 0 \, \text{V}) \]

**For Plate R:**
\[ V_{0,R} = 0.54 \, \text{V} \]

### Step 5: Analyze the I-V Characteristics
- For Plate P, we will have a stopping potential of \( 0.25 \, \text{V} \) and a current that saturates after reaching this potential.
- For Plate Q, there will be no current since the energy of the incident photons is less than the work function.
- For Plate R, the stopping potential is \( 0.54 \, \text{V} \), and it will also saturate at this value.

### Conclusion
The I-V graph will show:
- A current for Plate P that saturates at \( 0.25 \, \text{V} \).
- No current for Plate Q.
- A current for Plate R that saturates at \( 0.54 \, \text{V} \).

Thus, the correct I-V graph will show:
- A line for Plate P saturating at \( 0.25 \, \text{V} \).
- A flat line for Plate Q (no current).
- A line for Plate R saturating at \( 0.54 \, \text{V} \).