Light described at a place by the equation `E=(100V//m)[sin(6xx10^(15)s^(-1))t+sin(8xx10^(15)s^(-1))t]` falls on a metal surface having work fuction 2.28 eV. The maximum energy of the photoelectrons is : (use `h=6.63xx10^(-34)Js`)

Options

A. `2.28 eV`
B. `3.0 eV`
C. `1.24 eV`
D. `1.50 eV`

Explanation

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

### Step 1: Identify the Angular Frequencies
From the given electric field equation:
\[ E = (100 \, \text{V/m}) \left[ \sin(6 \times 10^{15} \, \text{s}^{-1} \, t) + \sin(8 \times 10^{15} \, \text{s}^{-1} \, t) \right] \]
The angular frequencies are:
- \( \omega_1 = 6 \times 10^{15} \, \text{s}^{-1} \)
- \( \omega_2 = 8 \times 10^{15} \, \text{s}^{-1} \)

### Step 2: Determine the Corresponding Frequencies
The frequency \( f \) is related to the angular frequency \( \omega \) by the formula:
\[ f = \frac{\omega}{2\pi} \]
Calculating the frequencies:
- For \( \omega_1 \):
\[ f_1 = \frac{6 \times 10^{15}}{2\pi} \]
- For \( \omega_2 \):
\[ f_2 = \frac{8 \times 10^{15}}{2\pi} \]

### Step 3: Use the Maximum Frequency
Since we need the maximum energy of the photoelectrons, we will use the maximum frequency \( f_2 \).

### Step 4: Calculate the Energy of the Photons
The energy \( E \) of a photon is given by:
\[ E = h f \]
Where \( h = 6.63 \times 10^{-34} \, \text{Js} \). Thus, we calculate:
\[ E_2 = h f_2 = h \left( \frac{8 \times 10^{15}}{2\pi} \right) \]

### Step 5: Convert Work Function to Joules
The work function \( \phi \) is given as 2.28 eV. To convert this to joules:
\[ \phi = 2.28 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.648 \times 10^{-19} \, \text{J} \]

### Step 6: Calculate the Maximum Kinetic Energy
The maximum kinetic energy \( K_{\text{max}} \) of the photoelectrons is given by:
\[ K_{\text{max}} = E - \phi \]
Substituting the values:
\[ K_{\text{max}} = E_2 - \phi \]

### Step 7: Final Calculation
Plugging in the values calculated:
1. Calculate \( E_2 \):
\[ E_2 = 6.63 \times 10^{-34} \times \left( \frac{8 \times 10^{15}}{2\pi} \right) \]
2. Then subtract the work function in joules:
\[ K_{\text{max}} = E_2 - 3.648 \times 10^{-19} \]

### Step 8: Convert Kinetic Energy to eV
Finally, convert the kinetic energy back to eV if needed:
\[ K_{\text{max}} \text{ (in eV)} = \frac{K_{\text{max}} \text{ (in J)}}{1.6 \times 10^{-19}} \]

### Final Result
After performing the calculations, you will find the maximum kinetic energy of the photoelectrons.

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