An electromagnetic radiation whose electric component varies with time as `E=E_0(1+cos omegat)cos omega_0t(omega=6xx10^(15) "rad"//s, omega_0=2.8 xx10^(14) "rad"//s)` is incident on a lithium surface (work function =2.39 eV) .
The maximum kinetic energy of a photoelectron liberated from the lithium surface is `(h=4.14 x10^(-15) ev-s)

Options

A. 2.50eV
B. 1.90eV
C. 1.75 eV
D. 1.50 eV

Answer

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

### Step 1: Understand the given electric field equation
The electric component of the electromagnetic radiation is given as:
\[ E = E_0(1 + \cos(\omega t)) \cos(\omega_0 t) \]
where:
- \( \omega = 6 \times 10^{15} \, \text{rad/s} \)
- \( \omega_0 = 2.8 \times 10^{14} \, \text{rad/s} \)

### Step 2: Simplify the electric field equation
Using the trigonometric identity for the product of cosines, we can rewrite the equation:
\[ E = E_0 \left( \cos(\omega_0 t) + \frac{1}{2} \cos((\omega + \omega_0)t) + \frac{1}{2} \cos((\omega - \omega_0)t) \right) \]

### Step 3: Identify the maximum frequency
The maximum frequency \( f_m \) can be derived from the maximum angular frequency:
\[ \omega_m = \omega + \omega_0 \]
Thus:
\[ f_m = \frac{\omega_m}{2\pi} = \frac{\omega + \omega_0}{2\pi} \]

Substituting the values:
\[ f_m = \frac{6 \times 10^{15} + 2.8 \times 10^{14}}{2\pi} \]

### Step 4: Calculate the maximum frequency
Calculating \( \omega + \omega_0 \):
\[ \omega + \omega_0 = 6 \times 10^{15} + 2.8 \times 10^{14} = 6.28 \times 10^{15} \, \text{rad/s} \]

Now, substituting into the frequency equation:
\[ f_m = \frac{6.28 \times 10^{15}}{2\pi} \approx 10^{15} \, \text{Hz} \]

### Step 5: Use the photoelectric effect equation
According to the photoelectric effect, the maximum kinetic energy \( K_{\text{max}} \) of the emitted photoelectron is given by:
\[ K_{\text{max}} = hf_m - \phi \]
where:
- \( h = 4.14 \times 10^{-15} \, \text{eV s} \) (Planck's constant)
- \( \phi = 2.39 \, \text{eV} \) (work function)

### Step 6: Calculate the maximum kinetic energy
Substituting the values into the equation:
\[ K_{\text{max}} = (4.14 \times 10^{-15} \, \text{eV s}) \times (10^{15} \, \text{Hz}) - 2.39 \, \text{eV} \]

Calculating:
\[ K_{\text{max}} = 4.14 - 2.39 = 1.75 \, \text{eV} \]

### Final Answer
The maximum kinetic energy of a photoelectron liberated from the lithium surface is:
\[ K_{\text{max}} = 1.75 \, \text{eV} \]

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