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 `3 lambda`, the stopping potential is `V_(0) .` The work function of the metallic surface is `(h c)/(k lambda)`. Find `k`.

Answer

To solve the problem, we will use the photoelectric effect equations and the information given about the stopping potentials when different wavelengths of light illuminate the metallic surface.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Equation**:
The photoelectric effect can be described by the equation:
\[ E_k = h \frac{c}{\lambda} - \phi \]
where \(E_k\) is the kinetic energy of the emitted electrons, \(h\) is Planck's constant, \(c\) is the speed of light, \(\lambda\) is the wavelength of the incident light, and \(\phi\) is the work function of the metal.

2. **Relating Stopping Potential to Kinetic Energy**:
The stopping potential \(V\) is related to the kinetic energy of the emitted electrons:
\[ E_k = eV \]
where \(e\) is the charge of the electron. Therefore, we can write:
\[ eV = h \frac{c}{\lambda} - \phi \]

3. **Setting Up the First Condition**:
For the first case with wavelength \(\lambda\) and stopping potential \(5V_0\):
\[ e(5V_0) = h \frac{c}{\lambda} - \phi \tag{1} \]

4. **Setting Up the Second Condition**:
For the second case with wavelength \(3\lambda\) and stopping potential \(V_0\):
\[ eV_0 = h \frac{c}{3\lambda} - \phi \tag{2} \]

5. **Expressing Work Function**:
We are given that the work function \(\phi\) is:
\[ \phi = \frac{hc}{k\lambda} \]

6. **Substituting Work Function into Equations**:
Substituting \(\phi\) into equations (1) and (2):
- From equation (1):
\[ e(5V_0) = h \frac{c}{\lambda} - \frac{hc}{k\lambda} \]
Simplifying gives:
\[ e(5V_0) = hc\left(\frac{1}{\lambda} - \frac{1}{k\lambda}\right) \]
\[ e(5V_0) = \frac{hc(k-1)}{k\lambda} \]

- From equation (2):
\[ eV_0 = h \frac{c}{3\lambda} - \frac{hc}{k\lambda} \]
Simplifying gives:
\[ eV_0 = hc\left(\frac{1}{3\lambda} - \frac{1}{k\lambda}\right) \]
\[ eV_0 = \frac{hc(k-3)}{3k\lambda} \]

7. **Setting Up the Ratio**:
Now we can set up a ratio of the two equations:
\[ \frac{e(5V_0)}{eV_0} = \frac{\frac{hc(k-1)}{k\lambda}}{\frac{hc(k-3)}{3k\lambda}} \]
This simplifies to:
\[ 5 = \frac{3(k-1)}{k-3} \]

8. **Solving for \(k\)**:
Cross-multiplying gives:
\[ 5(k-3) = 3(k-1) \]
Expanding both sides:
\[ 5k - 15 = 3k - 3 \]
Rearranging gives:
\[ 2k = 12 \implies k = 6 \]

### Final Answer:
Thus, the value of \(k\) is:
\[ \boxed{6} \]