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`.

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the number of incident photons per second
The intensity of the photon beam is given as \( I = 2.0 \, \text{W/m}^2 \) and the area of the surface is \( A = 1.0 \times 10^{-4} \, \text{m}^2 \). The energy of each photon is given as \( E_{\text{photon}} = 10.6 \, \text{eV} \).

First, we convert the energy of the photon from eV to Joules:
\[ E_{\text{photon}} = 10.6 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.696 \times 10^{-18} \, \text{J} \]

Now, we can calculate the total power incident on the surface:
\[ P = I \times A = 2.0 \, \text{W/m}^2 \times 1.0 \times 10^{-4} \, \text{m}^2 = 2.0 \times 10^{-4} \, \text{W} \]

Next, we find the number of photons incident per second:
\[ \text{Number of photons per second} = \frac{P}{E_{\text{photon}}} = \frac{2.0 \times 10^{-4} \, \text{W}}{1.696 \times 10^{-18} \, \text{J}} \approx 1.178 \times 10^{14} \, \text{photons/s} \]

### Step 2: Calculate the number of photoelectrons emitted per second
Given that 0.53% of the incident photons eject photoelectrons, we can calculate the number of photoelectrons emitted per second:
\[ \text{Number of photoelectrons emitted per second} = 0.0053 \times \text{Number of photons per second} \]
\[ \text{Number of photoelectrons emitted per second} = 0.0053 \times 1.178 \times 10^{14} \approx 6.25 \times 10^{11} \]

### Step 3: Calculate the minimum and maximum energy of the emitted photoelectrons
The work function \( \phi \) of platinum is given as \( 5.6 \, \text{eV} \).

1. **Minimum Kinetic Energy**: The minimum kinetic energy of the emitted photoelectrons can be considered as 0 eV if the photon energy is just equal to the work function.
\[ KE_{\text{min}} = 0 \, \text{eV} \]

2. **Maximum Kinetic Energy**: The maximum kinetic energy is given by the difference between the photon energy and the work function:
\[ KE_{\text{max}} = E_{\text{photon}} - \phi = 10.6 \, \text{eV} - 5.6 \, \text{eV} = 5.0 \, \text{eV} \]

### Final Answers:
- Number of photoelectrons emitted per second: \( 6.25 \times 10^{11} \)
- Minimum energy of emitted photoelectrons: \( 0 \, \text{eV} \)
- Maximum energy of emitted photoelectrons: \( 5.0 \, \text{eV} \)