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### Photo electric effect.

posted Oct 28, 2016, 12:40 PM by Upali Salpadoru   [ updated Jun 14, 2017, 8:59 PM ]

Fig.1   The Plan of his experiment. Fig.2. Heinrich Rudolf Hertz Ger. 1857 –1894

In 1887  Hertz noticed that when two oppositely charged metal rods were exposed to ultraviolet light , there was an electric spark. Wilhelm Hallwachs, one of his coleagues established the photoelectric effect using a very simple experiment.

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Fig. 3. Hallwachs experiment.

He used a gold leaf electroscope for this.

When a positively charged electroscope was exposed to bright light , it got discharged. When a negatively charged one was used, that did not happen.

In 1901 Max Planck had proposed that  electromagnetic energy may be emitted only in small lumps which he called quanta. This is somewhat similar to the existence of atoms in matter.

Fig.4 Philip Lenard 1862-1947.(Ger)

In  1902 Philip Lenard (Ger) . studied  how the kinetic energy of the emitted electrons varied with the intensity of light.

He used a very powerful light that can change the intensity. (brightness)  The emitted electrons could be prevented from reaching the  collector by giving the latter a negative charge. The necessary charge to oppose the electrons gave a measure of the energy in the emitted electrons.

Fig.5. Plan for the experiment of Leonard.

## Lenard discovered the following.

1. The particles emitted by metals on exposure to light were negatively charged. They were identical to the electrons discovered by JJ Thomson in …….’

2. Contrary to the belief at the time , he found that the kinetic energy in the emitted electrons does not depend on the intensity of light. A brighter light only increased the number of electrons emitted.

3. He discovered that the energy in the emitted electrons increased with the frequency of the rays used.

2. ## The gradient (slope) of the graph   is equal to dy/dx. That is h = 2.4 / 6 x10 14= 4 x10 -15.

3. Light having a frequency below 4 x1014 cannot remove electrons from this metal.

1. Work function is the minimum energy required to release an electron by photo electric effect.

2. This gives the formula   E  =  h x Frequency .    E = hf.

## This value of ‘h’ has  been determined by highly sophisticated experiments to be  equal to…………………….. 4.135667662x 10 -15. eVs.

(This  is the same as =6.62607 × 10−34  J.s.)