The electric field of certain radiation is given by the equation `E = 200 {sin (4pi xx 10^10)t + sin(4pi xx 10^15)t}`
falls in a metal surface having work function 2.0 eV. The maximum kinetic energy (in eV) of the photoelectrons is [Plank's constant `(h) = 6.63 xx 10^(-34)Js` and electron charge `e = 1.6 xx 10^(-19)C]`

Options

A. 3.3
B. 4.3
C. 5.3
D. 6.3

Answer

To solve the problem, we need to find the maximum kinetic energy of the photoelectrons emitted from a metal surface when exposed to a specific electric field radiation. Let's break down the solution step by step.

### Step 1: Identify the Electric Field Equation
The electric field is given by:
\[ E = 200 \sin(4\pi \times 10^{10} t) + \sin(4\pi \times 10^{15} t) \]

### Step 2: Determine the Frequencies
The electric field consists of two sine functions, each with a different angular frequency. The angular frequencies are:
- \( \omega_1 = 4\pi \times 10^{10} \)
- \( \omega_2 = 4\pi \times 10^{15} \)

To find the corresponding frequencies (\( \nu \)), we use the relation:
\[ \nu = \frac{\omega}{2\pi} \]

Calculating the frequencies:
1. For \( \omega_1 \):
\[ \nu_1 = \frac{4\pi \times 10^{10}}{2\pi} = 2 \times 10^{10} \, \text{Hz} \]

2. For \( \omega_2 \):
\[ \nu_2 = \frac{4\pi \times 10^{15}}{2\pi} = 2 \times 10^{15} \, \text{Hz} \]

### Step 3: Choose the Maximum Frequency
Since we want the maximum kinetic energy, we take the higher frequency:
\[ \nu = 2 \times 10^{15} \, \text{Hz} \]

### Step 4: Calculate the Energy of the Photons
The energy of a photon is given by:
\[ E = h\nu \]
Where \( h = 6.63 \times 10^{-34} \, \text{Js} \).

Calculating the energy:
\[ E = 6.63 \times 10^{-34} \times 2 \times 10^{15} \]
\[ E = 1.326 \times 10^{-18} \, \text{J} \]

### Step 5: Convert Work Function to Joules
The work function \( \phi \) is given as 2 eV. To convert this to joules:
\[ \phi = 2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.2 \times 10^{-19} \, \text{J} \]

### Step 6: Calculate Maximum Kinetic Energy
The maximum kinetic energy \( K.E. \) of the emitted photoelectrons is given by:
\[ K.E. = E - \phi \]
Substituting the values:
\[ K.E. = 1.326 \times 10^{-18} - 3.2 \times 10^{-19} \]
\[ K.E. = 1.006 \times 10^{-18} \, \text{J} \]

### Step 7: Convert Kinetic Energy to Electron Volts
To convert the kinetic energy from joules to electron volts:
\[ K.E. = \frac{1.006 \times 10^{-18}}{1.6 \times 10^{-19}} \]
\[ K.E. \approx 6.2875 \, \text{eV} \]

### Final Answer
The maximum kinetic energy of the photoelectrons is approximately:
\[ K.E. \approx 6.3 \, \text{eV} \]

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