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Question

The planck's constant is h and frequency of a photon is ν then write the formula for Einstein photo electric equation.

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Solution

12mV2max=h(ννo).

According to Plank's quantum theory, light is emitted from a source in the forms of bundles of energy called photons. Energy of each photon is E=hν.
Einstein made use of this theory to explain how photo electric emission takes place.
According to Einstein, when photons of energy E=hν fall on a metal surface, they transfer their energy to the electrons of metal. When the energy of photon is larger than the minimum energy required by the electrons to leave the metal surface, the emission of electrons take place instantaneously.
He proposed that an electron absorbs one whole photon or none. The chance that an electron may absorb more then one electron is negligible because the number of photons is much lower than the electron. After absorbing the photon, an electron either leaves the surface or dissipates its energy within the metal in such a short interval that it has almost no chance to absorb second photon. An increase in intensity of light source simply increases the number of photon and the number of photo electrons but no increase in the energy of photo electron. However, increase in frequency increases the energy of photons and photo electrons.
According to Einstein's explanation of photoelectric emission, a photon of energy E performs two operations:
1. Removes the electron from the surface of metal
2. Supplies some part of energy to move photo electron towards anode
Since minimum amount of energy to remove electron from a surface is equal to work function, we can write Einstein equation as:
Energy Supplied = Energy Consumed in ejecting an electron + maximum Kinetic energy of electron
E= work function +(KE)max
But, ϕo=hνo;E=hνand(KE)max=12mv2o
So, hν=hνo+12mv2o
h(ννo)=12mv2o
This is the Einstein's photoelectric equation.

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