What is the kinetic energy of the emitted electrons when cesium is exposed to uv rays of frequency 1.9×1015hz? express your answer in joules to three significant figures?

The energy of a photon of the incident UV rays is

where is the Planck constant, and is the frequency of the radiation.

When the photon hits the cesium, part of its energy is used to extract the photoelectron from the cesium (and this is called work function, ), while the rest is given to the photoelectron as kinetic energy K:


The work function for cesium is

Therefore, we can find the value of the kinetic energy of the electron by using these data into the previous equation:

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