When a beam of `10.6 eV` photons of intensity `2.0 W //m^2` falls on a platinum surface of area `1.0xx10^(-4) m^2` and work function `5.6 eV, 0.53%` of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per second and their minimum and maximum energy (in eV).
Take `1 eV = 1.6xx 10^(-19) J`.

Options

A. `6.25xx10^(8)`
B. `1.25xx10^(9)`
C. `1.25xx10^(6)`
D. `6.25xx10^(11)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the energy incident per second on the surface
The formula to calculate the power (energy per second) incident on the surface is given by:
\[ P = I \times A \]
Where:
- \( P \) is the power (in watts)
- \( I \) is the intensity (in watts per square meter)
- \( A \) is the area (in square meters)

Given:
- \( I = 2.0 \, \text{W/m}^2 \)
- \( A = 1.0 \times 10^{-4} \, \text{m}^2 \)

Calculating:
\[ P = 2.0 \, \text{W/m}^2 \times 1.0 \times 10^{-4} \, \text{m}^2 = 2.0 \times 10^{-4} \, \text{W} \]

### Step 2: Calculate the energy of each photon
The energy of each photon is given as:
\[ E_{\text{photon}} = 10.6 \, \text{eV} \]
To convert this to joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E_{\text{photon}} = 10.6 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.696 \times 10^{-18} \, \text{J} \]

### Step 3: Calculate the number of incident photons per second
The number of photons incident per second can be calculated using the formula:
\[ N_{\text{incident}} = \frac{P}{E_{\text{photon}}} \]
Substituting the values:
\[ N_{\text{incident}} = \frac{2.0 \times 10^{-4} \, \text{W}}{1.696 \times 10^{-18} \, \text{J}} \approx 1.176 \times 10^{14} \, \text{photons/s} \]

### Step 4: Calculate the number of photoelectrons emitted per second
Given that \( 0.53\% \) of the incident photons eject photoelectrons, we can calculate:
\[ N_{\text{emitted}} = 0.53\% \times N_{\text{incident}} = \frac{0.53}{100} \times 1.176 \times 10^{14} \approx 6.22 \times 10^{11} \, \text{photoelectrons/s} \]

### Step 5: Calculate the maximum kinetic energy of the emitted photoelectrons
Using Einstein's photoelectric equation:
\[ E_k = E_{\text{photon}} - \phi \]
Where \( \phi \) is the work function. Given:
- \( \phi = 5.6 \, \text{eV} \)

Calculating:
\[ E_k = 10.6 \, \text{eV} - 5.6 \, \text{eV} = 5.0 \, \text{eV} \]

### Step 6: Minimum energy of the emitted photoelectrons
The minimum energy of the emitted photoelectrons is equal to the kinetic energy when the photon energy is just sufficient to overcome the work function:
\[ E_{\text{min}} = 0 \, \text{eV} \quad (\text{when } E_{\text{photon}} = \phi) \]

### Summary of Results
- Number of photoelectrons emitted per second: \( \approx 6.22 \times 10^{11} \)
- Minimum energy of emitted photoelectrons: \( 0 \, \text{eV} \)
- Maximum energy of emitted photoelectrons: \( 5.0 \, \text{eV} \)