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 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 step by step, we will use the photoelectric effect equation and the given information to find the maximum kinetic energy of the photoelectrons.

### Step 1: Understand the given electric field equation
The electric field of the radiation is given by:
\[ E = 200 \sin(4\pi \times 10^{10} t) + \sin(4\pi \times 10^{15} t) \]

### Step 2: Identify the frequency of the radiation
From the electric field equation, we can see that the term with the highest frequency is \( \sin(4\pi \times 10^{15} t) \).

The angular frequency \( \omega \) is given by:
\[ \omega = 4\pi \times 10^{15} \, \text{rad/s} \]

To find the frequency \( \nu \):
\[ \nu = \frac{\omega}{2\pi} = \frac{4\pi \times 10^{15}}{2\pi} = 2 \times 10^{15} \, \text{Hz} \]

### Step 3: Use the photoelectric effect equation
The maximum kinetic energy \( K_{\text{max}} \) of the emitted photoelectrons can be calculated using the equation:
\[ K_{\text{max}} = h\nu - W \]
where:
- \( h \) is Planck's constant \( (6.63 \times 10^{-34} \, \text{Js}) \)
- \( W \) is the work function in Joules.

### Step 4: Convert the work function from eV to Joules
The work function \( W \) is given as 2 eV. To convert this to Joules:
\[ W = 2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.2 \times 10^{-19} \, \text{J} \]

### Step 5: Calculate the maximum kinetic energy
Now we can substitute the values into the kinetic energy equation:
\[ K_{\text{max}} = h\nu - W \]
\[ K_{\text{max}} = (6.63 \times 10^{-34} \, \text{Js}) \times (2 \times 10^{15} \, \text{Hz}) - (3.2 \times 10^{-19} \, \text{J}) \]

Calculating \( h\nu \):
\[ h\nu = 6.63 \times 10^{-34} \times 2 \times 10^{15} = 1.326 \times 10^{-18} \, \text{J} \]

Now substituting back into the equation:
\[ K_{\text{max}} = 1.326 \times 10^{-18} - 3.2 \times 10^{-19} \]
\[ K_{\text{max}} = 1.006 \times 10^{-18} \, \text{J} \]

### Step 6: Convert the maximum kinetic energy back to eV
To convert Joules back to electron volts:
\[ K_{\text{max}} = \frac{1.006 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 6.29 \, \text{eV} \]

### Final Answer
The maximum kinetic energy of the photoelectrons is approximately \( 6.3 \, \text{eV} \).

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