C

Step 1: According to the photoelectric effect, the stopping potential (V) is related to the energy of the incident photons. The energy of a photon is given by the equation: $E = \frac{hc}{\lambda}$ where: - $E$ is the energy of the photon, - $h$ is Planck's constant, - $c$ is the speed of light, - $\lambda$ is the wavelength of the radiation.

Step 2: For the first case with wavelength $\lambda$, the stopping potential is given by: $eV = \frac{hc}{\lambda} - \phi$ where $\phi$ is the work function of the metal.

Step 3: For the second case with wavelength $3\lambda$, the stopping potential is: $eV' = \frac{hc}{3\lambda} - \phi$

Step 4: The stopping potential $V'$ can be expressed in relation to $V$: $V' = \frac{V}{3} + \frac{\phi}{e} - \frac{\phi}{e} = \frac{V}{3}$ (rearranging to find the new stopping potential).

Step 5: Given that the stopping potential for radiation of wavelength $3\lambda$ is $\frac{V}{4}$, we can set up the following equation: $\frac{V}{3} = \frac{V}{4} + \frac{\phi}{e}$

Step 6: Solve for $\phi$: Multiplying through by 12: $4V = 3V + 12\frac{\phi}{e}$
Solving gives: $\frac{\phi}{e} = \frac{V}{12}$

Step 7: The threshold wavelength $\lambda_0$ satisfies: $\phi = \frac{hc}{\lambda_0}$.
From the above, we can substitute: $\frac{hc}{\lambda_0} = \frac{V}{12}$.
This implies: $\lambda_0 = \frac{12hc}{V}$.
Substituting back in terms of $n \lambda$: Using the stopping potential we can express: $\lambda_0 = n \lambda$, leading to $n = 12$.
Therefore, the value of n will be: 12.