What is the kinetic energy of the emitted electrons when cesium is exposed to uv rays of frequency 1.0×1015hz?

The kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.0x10^15Hz is 3.266x10^-19J.

Further explanation

Calculations

As   Energy=hf

E=6.626x10^-34 x 1.0x10^15

E=6.626x10^-19J

Since work function of cesium=2.1eV

And by converting its value in joule unit it will be

2.1eV=3.36x10^-19

KE of the emitted electrons=hf-(work function)

KE=6.626x10^-19J - 3.36x10^-19

KE=3.266x10^-19J

Explanation

The photon of light knocks out electrons from any metal surface. One photon emits one electron from metal surface. A photon must have enough energy to knock out electrons from a metal surface because if the energy of the incident photon will be less than the work function of metal surface than no electrons will be ejected from the metal surface. The kinetic energy of emitted electrons is the energy left after knocking out electron as some energy is used to knock out electrons. So the kinetic energy of emitted electrons is equal to the total energy of photon minus the work function of metal.  

Work function  

Work function is the minimum amount of energy required to eject an electron from a metal surface. The number of electrons which are emitted and their kinetic energy can be measured as function of frequency of light and its intensity. The kinetic energy of the emitted electrons is equal to the energy of the incident photon minus the work function of metal.  

Answer details

Subject: Chemistry

Level: College

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i think its 100

i hope this !