The energy of the incident photon $E=\frac{hc}{\lambda }$will decrease if we increase the wavelength $\left(\lambda \right)$ of the incident light.

Here $h$ is Planck constant and $c$ is the speed of light in vacuum.

As a result, the energy released by the photoelectron reduces, but the number of emitted photoelectrons does not change.

According to Einstein's photoelectric equation, $K.E_{\max }=\frac{hc}{\lambda }-{\varphi }_0$, here ${\varphi }_0$ is the work function. 

As seen from the above equation, kinetic energy and wavelength are inversely proportional to each other, Thus, with an increase in wavelength of the incident light, the kinetic energy of photoelectrons decreases.