The answer to this question lies in the relationship between the frequency of light and the energy of photons. In a photoelectric experiment, when light shines on a metal surface, electrons can be ejected from the surface if the energy of the photons is sufficient to overcome the work function of the metal.

1. Relationship between frequency and energy:
The energy of a photon is directly proportional to its frequency. This relationship is given by the equation E = hf, where E is the energy of the photon, h is Planck's constant, and f is the frequency of the light. Thus, if the frequency of light is doubled, the energy of the photons will also double.

2. Stopping potential and energy of electrons:
In a photoelectric experiment, once the electrons are ejected from the metal surface, they experience a potential difference (stopping potential) that opposes their motion. The stopping potential is the minimum potential required to stop the electrons from reaching a detector.

The stopping potential is directly related to the maximum kinetic energy of the ejected electrons. The maximum kinetic energy of an ejected electron is given by the difference between the energy of the incident photon and the work function of the metal.

3. Effect of doubling the frequency:
When the frequency of light is doubled, the energy of the photons is also doubled. This means that the maximum kinetic energy of the ejected electrons will also increase. As a result, the stopping potential required to stop these high-energy electrons will be greater than double the original value.

Therefore, the correct answer is option 'C' - the stopping potential will become more than doubled when the frequency of light is doubled in a photoelectric experiment.

In summary, doubling the frequency of light in a photoelectric experiment increases the energy of photons, which in turn increases the maximum kinetic energy of ejected electrons. This requires a greater stopping potential to stop the high-energy electrons from reaching a detector.