In a photoelectric experiment, the collector plate is at 2.0V with respect to the emitter plate made of copper `(phi)=4.5eV)`. The emitter is illuminated by a source of monochromatic light of wavelength 2000 `Å`nm.

Options

A. The maximum kinetic energy of the photoelctrons reaching the collector is `0`.
B. The maximum kinetic energy of the photoelctrons reaching the collector is `3.7` .eV.
C. If the polarity of the battery is reversed then answer to part A willl be 0
D. If the polarity of the battery is reversed then answer to part B will be `1.7 eV`

Explanation

To solve the problem step by step, we will use the photoelectric effect equation and the given parameters.

### Step 1: Understand the given parameters
- Work function of copper, \( \phi = 4.5 \, \text{eV} \)
- Collector plate voltage, \( V_0 = 2.0 \, \text{V} \)
- Wavelength of light, \( \lambda = 2000 \, \text{Å} = 2000 \times 10^{-10} \, \text{m} \)

### Step 2: Calculate the energy of the incident photons
We can calculate the energy of the incident photons using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h = 6.63 \times 10^{-34} \, \text{Js} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)

Substituting the values:
\[ E = \frac{(6.63 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{2000 \times 10^{-10} \, \text{m}} \]

### Step 3: Perform the calculation
Calculating the numerator:
\[ 6.63 \times 10^{-34} \times 3 \times 10^8 = 1.989 \times 10^{-25} \, \text{Jm} \]
Now, divide by the wavelength:
\[ E = \frac{1.989 \times 10^{-25}}{2000 \times 10^{-10}} = \frac{1.989 \times 10^{-25}}{2 \times 10^{-7}} = 9.945 \times 10^{-19} \, \text{J} \]

### Step 4: Convert energy from Joules to electron volts
To convert Joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = \frac{9.945 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 6.215625 \, \text{eV} \]

### Step 5: Calculate the maximum kinetic energy of emitted electrons
Using the photoelectric equation:
\[ K_{\text{max}} = E - \phi + eV_0 \]
Substituting the values:
\[ K_{\text{max}} = 6.215625 \, \text{eV} - 4.5 \, \text{eV} + 2.0 \, \text{V} \]
\[ K_{\text{max}} = 6.215625 - 4.5 + 2 = 3.715625 \, \text{eV} \]

### Final Answer
The maximum kinetic energy of the emitted electrons is approximately \( 3.72 \, \text{eV} \).

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