What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.0*1015Hz?

In order to emit electrons, the cesium will have to absorb photons. Each photon will knock out one electron by transferring its energy to the electron. Therefore, by the principle of energy conservation, the energy of the removed electron will be equal to the energy of the incident photon. That energy is calculated using Planck's equation:

E = hf

E = 6.63 x 10⁻³⁴ * 1 x 10¹⁵

E = 6.63 x 10⁻¹⁹ Joules

The electron will have 6.63 x 10⁻¹⁹ Joules of kinetic energy