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Question

Explain an experiment to demonstrate photoelectric effect.

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Solution

It consists of a tube having two electrodes in vacuum . The electrode also called cathode is a photosensitive emitter which emits photoelectrons when exposed to light. The electrode A is a charge collecting plate. The tube has a side window, made of quartz covered with a filter through which the incident light odf desired frequency enters and falls on photosensitive plate c. Electrodes A and C are connected to a battery through a reversible switch S. Electrode A can be bought to a positive or a negative potential heat electrode C with help of this switch.

Electrons are emitted when the light is made to fall on plate ‘C’. These electrons get attracted towards A. When it is positive potential at C . This flow of electrons from C to A makes electric current to flow in the circuit. This current is called as photoelectric current as is due to photoelectrons emitted by photosensitive cathode ‘C’. This current is measured with microammeter connected in series. The potential difference between A and C is measured with voltmeter V.


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