Step 1: Recall the condition mentioned in the question.
The light frequency is already above the threshold frequency. This means photoelectric emission is taking place. So electrons are being emitted from the photosensitive surface.
Step 2: Recall Einstein's photoelectric equation.
The maximum kinetic energy of emitted electrons is given by:
\[
K_{\max}=h\nu-\phi
\]
where \(h\nu\) is the energy of the incident photon and \(\phi\) is the work function of the material.
Step 3: Identify what changes and what remains fixed.
In this question, only the intensity of incident light is increased. The frequency \(\nu\) remains unchanged. Therefore, the energy of each photon remains the same.
Step 4: Determine the effect on maximum kinetic energy and stopping potential.
Since \(K_{\max}\) depends only on the frequency of light and not on its intensity, increasing intensity does not increase the maximum kinetic energy. Therefore, the stopping potential, which is directly related to \(K_{\max}\), also does not increase.
Step 5: Interpret intensity in terms of photons.
Intensity of light is related to the number of photons incident per unit time. If intensity increases while frequency remains the same, then more photons fall on the surface every second.
Step 6: Relate number of photons to photoelectrons emitted.
Since each photon can liberate at most one electron, more incident photons mean more emitted electrons, provided the frequency is above threshold. Hence, photoelectric current increases.
Step 7: State the final answer.
Therefore, when the intensity is increased above threshold frequency, the correct statement is:
\[
\boxed{\text{The number of emitted electrons increases}}
\]
which matches option \((1)\).