In an experiment on the photoelectric effect, if the wavelength of...

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by K_\max = \dfrac{hc}{\lambda} - \Phi, where $\Phi$ is the work function of the metal and $V_s$ is the stopping potential such that K_\max = eV_s.

When the wavelength of the incident light is doubled, the new energy of the incident photon becomes $E' = \dfrac{hc}{2\lambda} = \dfrac{E}{2}$.

If $\dfrac{E}{2} < \Phi$, photoelectric emission will stop. If $\dfrac{E}{2} > \Phi$, photoelectric emission will continue. Thus, photoelectric emission may or may not occur.

If emission continues, the new stopping potential $V_s'$ is given by $eV_s' = \dfrac{E}{2} - \Phi = \dfrac{\Phi + eV_s}{2} - \Phi = \dfrac{eV_s - \Phi}{2}$.

Since $\Phi > 0$, we have $V_s' < \dfrac{V_s}{2} < V_s$. Therefore, the stopping potential will decrease.