A photoelectric surface is illuminated successively by monochromatic light of wavelength `lambda` and `(lambda)/(2)`. If the maximum kinetic energy of the emitted photoelectrons in the second case is `3` times than in the first case , the work function of the surface of the material is
(`h` = Plank's constant , `c` = speed of light )

Options

A. `(hc)/(2lambda)`
B. `(hc)/(lambda)`
C. `(2hc)/(lambda)`
D. `(hc)/(3lambda)`

Answer

To solve the problem, we need to analyze the photoelectric effect using the given information about the wavelengths and the kinetic energies of the emitted photoelectrons. Let's break it down step by step.

### Step 1: Understand the photoelectric effect equation
The maximum kinetic energy (K.E.) of the emitted photoelectrons can be expressed by the equation:
\[ K.E. = E - \Phi \]
where \( E \) is the energy of the incident photons and \( \Phi \) is the work function of the material.

### Step 2: Calculate the energy of the photons for the first case
For the first case, where the wavelength is \( \lambda \):
\[ E_1 = \frac{hc}{\lambda} \]
Thus, the maximum kinetic energy \( K_1 \) can be expressed as:
\[ K_1 = E_1 - \Phi = \frac{hc}{\lambda} - \Phi \]

### Step 3: Calculate the energy of the photons for the second case
In the second case, the wavelength is \( \frac{\lambda}{2} \):
\[ E_2 = \frac{hc}{\frac{\lambda}{2}} = \frac{2hc}{\lambda} \]
The maximum kinetic energy \( K_2 \) for this case is:
\[ K_2 = E_2 - \Phi = \frac{2hc}{\lambda} - \Phi \]

### Step 4: Relate the kinetic energies from the problem statement
According to the problem, the maximum kinetic energy in the second case is three times that of the first case:
\[ K_2 = 3K_1 \]

### Step 5: Substitute the expressions for kinetic energies
Substituting the expressions for \( K_1 \) and \( K_2 \):
\[ \frac{2hc}{\lambda} - \Phi = 3\left(\frac{hc}{\lambda} - \Phi\right) \]

### Step 6: Expand and simplify the equation
Expanding the right side:
\[ \frac{2hc}{\lambda} - \Phi = \frac{3hc}{\lambda} - 3\Phi \]

Now, rearranging gives:
\[ \frac{2hc}{\lambda} - \frac{3hc}{\lambda} = -3\Phi + \Phi \]
\[ -\frac{hc}{\lambda} = -2\Phi \]

### Step 7: Solve for the work function \( \Phi \)
Dividing both sides by -1:
\[ \frac{hc}{\lambda} = 2\Phi \]
Thus, we can express the work function as:
\[ \Phi = \frac{hc}{2\lambda} \]

### Final Answer
The work function of the surface of the material is:
\[ \Phi = \frac{hc}{2\lambda} \]

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