Electromagnetic radiation whose electric component varies sinusoidally with time as `E=C_1(C_2+C_3 cos omegat) cos omega_0t , "where" C_1, C_2 and C_3` are constants in incident on lithium and liberates photoelectrons. If the kinetic energy of the most energetic photoelectrons be `0.592 xx10^(-19)`J find the work function of lithium given that `omega_0= 3.6 xx10^(15)` rad/sec and `omega=6 xx10^(14)` rad/sec.

Answer

To solve the problem, we need to find the work function (φ) of lithium using the given information about the electromagnetic radiation and the kinetic energy of the photoelectrons. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Kinetic energy of the most energetic photoelectrons, \( K.E. = 0.592 \times 10^{-19} \, \text{J} \)
- Angular frequency of the radiation, \( \omega_0 = 3.6 \times 10^{15} \, \text{rad/s} \)
- Angular frequency of the oscillating electric field, \( \omega = 6 \times 10^{14} \, \text{rad/s} \)

### Step 2: Calculate the maximum frequency (ν_max)
The maximum frequency (ν_max) is given by:
\[ \nu_{\text{max}} = \frac{\omega + \omega_0}{2\pi} \]
Substituting the values:
\[ \nu_{\text{max}} = \frac{(6 \times 10^{14} + 3.6 \times 10^{15})}{2\pi} \]
Calculating the sum:
\[ \nu_{\text{max}} = \frac{(4.2 \times 10^{15})}{2\pi} \]
Now, using \( \pi \approx 3.14 \):
\[ \nu_{\text{max}} \approx \frac{4.2 \times 10^{15}}{6.28} \approx 6.69 \times 10^{14} \, \text{Hz} \]

### Step 3: Use the photoelectric equation
The photoelectric equation is given by:
\[ K.E. = h\nu_{\text{max}} - \phi \]
Where:
- \( h \) is Planck's constant, \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- \( \phi \) is the work function

Rearranging the equation to find the work function:
\[ \phi = h\nu_{\text{max}} - K.E. \]

### Step 4: Calculate \( h\nu_{\text{max}} \)
First, calculate \( h\nu_{\text{max}} \):
\[ h\nu_{\text{max}} = h \cdot (6.69 \times 10^{14}) = 6.626 \times 10^{-34} \times 6.69 \times 10^{14} \]
Calculating this gives:
\[ h\nu_{\text{max}} \approx 4.44 \times 10^{-19} \, \text{J} \]

### Step 5: Substitute into the work function equation
Now substitute \( h\nu_{\text{max}} \) and \( K.E. \) into the work function equation:
\[ \phi = 4.44 \times 10^{-19} - 0.592 \times 10^{-19} \]
Calculating this gives:
\[ \phi \approx 3.848 \times 10^{-19} \, \text{J} \]

### Step 6: Convert the work function to electron volts
To convert joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ \phi_{\text{eV}} = \frac{3.848 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.40 \, \text{eV} \]

### Final Answer
The work function of lithium is approximately \( \phi \approx 2.40 \, \text{eV} \).

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