When a metallic surface is illuminated with radiation of wavelength `lambda`, the stopping potential is `V`. If the same surface is illuminated with radiation of wavelength `2 lambda` , the stopping potential is `(V)/(4)`. The threshold wavelength surface is :
(a) `5lambda` (b) `(5)/(2)lambda` (c) `3lambda` (d) `4lambda`

Options

A. `5lambda`
B. `(5)/(2)lambda`
C. `3lambda`
D. `4lambda`

Answer

To solve the problem, we will use the principles of the photoelectric effect and the equations derived from it. Here’s a step-by-step breakdown:

### Step 1: Understand the Photoelectric Effect
The photoelectric effect can be described by the equation:
\[ E = eV_0 = E_{\text{incident}} - \phi \]
where:
- \( E \) is the energy of the emitted electrons,
- \( e \) is the charge of the electron,
- \( V_0 \) is the stopping potential,
- \( E_{\text{incident}} \) is the energy of the incident photons,
- \( \phi \) is the work function of the material.

### Step 2: Write the Energy of Incident Photons
The energy of the incident photons can be expressed as:
\[ E_{\text{incident}} = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident radiation.

### Step 3: Set Up the Equations for Two Cases
1. For the first case with wavelength \( \lambda \) and stopping potential \( V \):
\[ eV = \frac{hc}{\lambda} - \phi \]
Rearranging gives us:
\[ eV + \phi = \frac{hc}{\lambda} \tag{1} \]

2. For the second case with wavelength \( 2\lambda \) and stopping potential \( \frac{V}{4} \):
\[ e\left(\frac{V}{4}\right) = \frac{hc}{2\lambda} - \phi \]
Rearranging gives us:
\[ \frac{eV}{4} + \phi = \frac{hc}{2\lambda} \tag{2} \]

### Step 4: Equate the Work Function
Since the work function \( \phi \) is the same for both cases, we can set the two equations for \( \phi \) equal to each other:
From (1):
\[ \phi = \frac{hc}{\lambda} - eV \]
From (2):
\[ \phi = \frac{hc}{2\lambda} - \frac{eV}{4} \]

### Step 5: Set the Equations Equal
Setting the two expressions for \( \phi \) equal:
\[ \frac{hc}{\lambda} - eV = \frac{hc}{2\lambda} - \frac{eV}{4} \]

### Step 6: Solve for \( \lambda_0 \)
Rearranging the equation:
\[ \frac{hc}{\lambda} - \frac{hc}{2\lambda} = eV - \frac{eV}{4} \]
This simplifies to:
\[ \frac{hc}{2\lambda} = \frac{3eV}{4} \]
Cross-multiplying gives:
\[ 4hc = 6\lambda eV \]
Thus:
\[ \lambda = \frac{2hc}{3eV} \]

### Step 7: Find the Threshold Wavelength
The threshold wavelength \( \lambda_0 \) is related to the work function by:
\[ \phi = \frac{hc}{\lambda_0} \]
Substituting \( \phi \) from either equation gives:
\[ \lambda_0 = \frac{hc}{\phi} \]
From the equations, we can derive that:
\[ \lambda_0 = 4\lambda \]

### Final Answer
Thus, the threshold wavelength of the surface is:
\[ \lambda_0 = 4\lambda \]
So the correct option is (d) \( 4\lambda \).

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