Photoelectric effect experiments are performed using three different metal plates `p,q` and`r` having work function `phi_(p) = 2.0 eV, phi_(e) = 2.5 eV and phi_(r) = 3.0 eV` respectively A light beam containing wavelength of `550nm , 450 nm` and `350nm ` with equal intensities illuminates each of the plates . The correct `I -V` graph for the experiment is [Take hc = 1240 eV nm]

Options

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B.
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D.

Explanation

To solve the problem related to the photoelectric effect experiments performed using three different metal plates (P, Q, and R) with given work functions and illuminated by light of different wavelengths, we will follow these steps:

### Step 1: Calculate the Energy of the Incident Photons
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the light.

Given:
- \( hc = 1240 \, \text{eV nm} \)

We will calculate the energy for each wavelength:

1. **For \( \lambda = 550 \, \text{nm} \)**:
\[ E_1 = \frac{1240 \, \text{eV nm}}{550 \, \text{nm}} \approx 2.25 \, \text{eV} \]

2. **For \( \lambda = 450 \, \text{nm} \)**:
\[ E_2 = \frac{1240 \, \text{eV nm}}{450 \, \text{nm}} \approx 2.76 \, \text{eV} \]

3. **For \( \lambda = 350 \, \text{nm} \)**:
\[ E_3 = \frac{1240 \, \text{eV nm}}{350 \, \text{nm}} \approx 3.54 \, \text{eV} \]

### Step 2: Compare Photon Energy with Work Functions
Now we compare the calculated photon energies with the work functions of the metals:

- **Metal P**: \( \phi_p = 2.0 \, \text{eV} \)
- **Metal Q**: \( \phi_q = 2.5 \, \text{eV} \)
- **Metal R**: \( \phi_r = 3.0 \, \text{eV} \)

1. **For Metal P**:
- \( E_1 = 2.25 \, \text{eV} \) (greater than \( \phi_p \))
- \( E_2 = 2.76 \, \text{eV} \) (greater than \( \phi_p \))
- \( E_3 = 3.54 \, \text{eV} \) (greater than \( \phi_p \))
- **Conclusion**: All wavelengths can eject electrons.

2. **For Metal Q**:
- \( E_1 = 2.25 \, \text{eV} \) (less than \( \phi_q \))
- \( E_2 = 2.76 \, \text{eV} \) (greater than \( \phi_q \))
- \( E_3 = 3.54 \, \text{eV} \) (greater than \( \phi_q \))
- **Conclusion**: Only \( E_2 \) and \( E_3 \) can eject electrons.

3. **For Metal R**:
- \( E_1 = 2.25 \, \text{eV} \) (less than \( \phi_r \))
- \( E_2 = 2.76 \, \text{eV} \) (less than \( \phi_r \))
- \( E_3 = 3.54 \, \text{eV} \) (greater than \( \phi_r \))
- **Conclusion**: Only \( E_3 \) can eject electrons.

### Step 3: Determine Stopping Potential
The stopping potential \( V_0 \) can be calculated using:
\[ V_0 = \frac{E - \phi}{e} \]
where \( e \) is the charge of an electron (1 eV corresponds to 1 V for the stopping potential).

1. **For Metal P**:
- Maximum energy of emitted electrons: \( E_3 - \phi_p = 3.54 - 2.0 = 1.54 \, \text{eV} \)

2. **For Metal Q**:
- Maximum energy of emitted electrons: \( E_2 - \phi_q = 2.76 - 2.5 = 0.26 \, \text{eV} \)

3. **For Metal R**:
- Maximum energy of emitted electrons: \( E_3 - \phi_r = 3.54 - 3.0 = 0.54 \, \text{eV} \)

### Step 4: Analyze the Current
The current in the photoelectric effect is proportional to the number of emitted electrons. Since:
- Metal P can emit electrons from all three wavelengths.
- Metal Q can emit from two wavelengths.
- Metal R can emit from one wavelength.

Thus, the order of current will be:
- **Metal P > Metal Q > Metal R**

### Conclusion
The correct \( I-V \) graph will show the highest current for Metal P, followed by Metal Q, and the least for Metal R.