When a metallic surface is illuminated with monochromatic light of wavelength `lambda`, the stopping potential is `5 V_0`. When the same surface is illuminated with light of wavelength `3lambda`, the stopping potential is `V_0`. Then the work function of the metallic surface is:

Options

A. `(hc)/(6lambda)`
B. `(hc)/(5lambda)`
C. `(hc)/(3lambda)`
D. `(hc)/(4lambda)`

Answer

To find the work function (φ) of the metallic surface based on the given stopping potentials and wavelengths, we can follow these steps:

### Step 1: Write down the photoelectric equation
The photoelectric equation relates the energy of the incident photons to the work function and the stopping potential:
\[ E = eV_0 + \phi \]
where \(E\) is the energy of the incident light, \(e\) is the charge of the electron, \(V_0\) is the stopping potential, and \(\phi\) is the work function.

### Step 2: Express the energy of the incident light
The energy of a photon can be expressed in terms of its wavelength (\(\lambda\)):
\[ E = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant and \(c\) is the speed of light.

### Step 3: Set up the equations for both wavelengths
1. For the first wavelength \(\lambda\) with stopping potential \(5V_0\):
\[ \frac{hc}{\lambda} = e(5V_0) + \phi \quad \text{(1)} \]

2. For the second wavelength \(3\lambda\) with stopping potential \(V_0\):
\[ \frac{hc}{3\lambda} = eV_0 + \phi \quad \text{(2)} \]

### Step 4: Rearrange both equations
From equation (1):
\[ \phi = \frac{hc}{\lambda} - e(5V_0) \quad \text{(3)} \]

From equation (2):
\[ \phi = \frac{hc}{3\lambda} - eV_0 \quad \text{(4)} \]

### Step 5: Set equations (3) and (4) equal to each other
Since both equations equal \(\phi\), we can set them equal:
\[ \frac{hc}{\lambda} - e(5V_0) = \frac{hc}{3\lambda} - eV_0 \]

### Step 6: Solve for \(V_0\)
Rearranging gives:
\[ \frac{hc}{\lambda} - \frac{hc}{3\lambda} = e(5V_0) - eV_0 \]
Factoring out common terms:
\[ \frac{hc}{\lambda} \left(1 - \frac{1}{3}\right) = e(5V_0 - V_0) \]
This simplifies to:
\[ \frac{hc}{\lambda} \cdot \frac{2}{3} = e(4V_0) \]

### Step 7: Solve for \(V_0\)
Now, solve for \(V_0\):
\[ V_0 = \frac{hc}{8e\lambda} \]

### Step 8: Substitute \(V_0\) back to find \(\phi\)
Substituting \(V_0\) back into equation (4):
\[ \phi = \frac{hc}{3\lambda} - e\left(\frac{hc}{8e\lambda}\right) \]
This simplifies to:
\[ \phi = \frac{hc}{3\lambda} - \frac{hc}{8\lambda} \]
Finding a common denominator (24):
\[ \phi = \frac{8hc}{24\lambda} - \frac{3hc}{24\lambda} = \frac{5hc}{24\lambda} \]

### Step 9: Final expression for work function
Thus, the work function \(\phi\) is:
\[ \phi = \frac{hc}{6\lambda} \]

### Conclusion
The work function of the metallic surface is:
\[ \phi = \frac{hc}{6\lambda} \]