Light wave described by the equation `200 V//m sin (1.5xx10^15 s^(-1) t cos (0.5xx10^15 s^(-1) t` falls metal surface having work function 2.0 eV. Then, the maximum kinetic energy photoelectrons is

Options

A. `3.27eV`
B. `2.2eV`
C. `2.85eV`
D. none of these

Explanation

To solve the problem, we need to find the maximum kinetic energy of photoelectrons emitted when a light wave falls on a metal surface with a given work function. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the given parameters
The light wave is described by the equation:
\[ E = 200 \, \text{V/m} \sin(1.5 \times 10^{15} \, \text{s}^{-1} \, t) \cos(0.5 \times 10^{15} \, \text{s}^{-1} \, t) \]
The work function (\( \phi \)) of the metal is given as \( 2.0 \, \text{eV} \).

### Step 2: Determine the frequency of the light wave
The angular frequency (\( \omega \)) of the light wave can be determined from the sine term in the equation. Here, the maximum angular frequency is:
\[ \omega_1 = 1.5 \times 10^{15} \, \text{s}^{-1} \]

To find the frequency (\( \nu \)), we use the relationship:
\[ \nu = \frac{\omega}{2\pi} \]
Calculating this gives:
\[ \nu = \frac{1.5 \times 10^{15}}{2\pi} \approx \frac{1.5 \times 10^{15}}{6.28} \approx 2.39 \times 10^{14} \, \text{Hz} \]

### Step 3: Calculate the energy of the photons
The energy (\( E \)) of the photons can be calculated using the formula:
\[ E = h \nu \]
Where \( h \) (Planck's constant) is approximately \( 6.63 \times 10^{-34} \, \text{Js} \). Thus:
\[ E = 6.63 \times 10^{-34} \times 2.39 \times 10^{14} \]

Calculating this gives:
\[ E \approx 1.58 \times 10^{-19} \, \text{J} \]

### Step 4: Convert energy from Joules to electron volts
To convert the energy from Joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E \approx \frac{1.58 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 0.9875 \, \text{eV} \]

### Step 5: Calculate the maximum kinetic energy of photoelectrons
The maximum kinetic energy (\( KE_{\text{max}} \)) of the emitted photoelectrons is given by:
\[ KE_{\text{max}} = E - \phi \]
Substituting the values we have:
\[ KE_{\text{max}} = 0.9875 \, \text{eV} - 2.0 \, \text{eV} \]
\[ KE_{\text{max}} = -1.0125 \, \text{eV} \]

### Step 6: Conclusion
Since the maximum kinetic energy is negative, this indicates that the energy of the incoming photons is less than the work function of the metal. Therefore, no photoemission occurs.

### Final Answer
The maximum kinetic energy of the photoelectrons is:
**None, as no photoemission takes place.**

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