
How does photoelectric effect prove light is made of particles?
Answer
544.2k+ views
Hint: The photoelectric effect follows the light theory of particles in that it functions as an elastic collision (one that preserves mechanical energy) between two particles, the light photon, and the metal electron.
Complete answer:
The photoelectric effect occurs as light reaches a metal surface, causing electrons to be released from it (photoelectrons). As a result, more electrons are released if you increase the strength of light, but their kinetic energy does not increase. The number of photoelectrons emitted does not increase as you increase the frequency of the incident light as the speed, and so their kinetic energy, increases... the electrons emitted are more... powerful!
In order to understand this, let us see an example: No electrons from the metal will be expelled if you shine a light on the metal of some strength with energy below the electron's binding energy. The electron from the metal will be knocked off the metal as soon as the light level is high enough so that the radiation reaches the binding energy.
If $hv$ is the energy of the photon that reaches the metal, so the energy in the collision can be conserved such that,
$hv = B.E + K.{E_{electron}}$
Where, $hv$ is the energy before the collision,$B.E$ is the binding energy which is the minimum amount of energy needed to eject the electron, $K.{E_{electron}}$ is the kinetic energy.
Note: The light consists of photons. What it means that the energy in the wave arrives in lumps as it is absorbed or expelled. The scale of such energy lumps (or 'quanta') depends on the wavelength. The larger the wavelength, the greater the sum of energy per photon.
Complete answer:
The photoelectric effect occurs as light reaches a metal surface, causing electrons to be released from it (photoelectrons). As a result, more electrons are released if you increase the strength of light, but their kinetic energy does not increase. The number of photoelectrons emitted does not increase as you increase the frequency of the incident light as the speed, and so their kinetic energy, increases... the electrons emitted are more... powerful!
In order to understand this, let us see an example: No electrons from the metal will be expelled if you shine a light on the metal of some strength with energy below the electron's binding energy. The electron from the metal will be knocked off the metal as soon as the light level is high enough so that the radiation reaches the binding energy.
If $hv$ is the energy of the photon that reaches the metal, so the energy in the collision can be conserved such that,
$hv = B.E + K.{E_{electron}}$
Where, $hv$ is the energy before the collision,$B.E$ is the binding energy which is the minimum amount of energy needed to eject the electron, $K.{E_{electron}}$ is the kinetic energy.
Note: The light consists of photons. What it means that the energy in the wave arrives in lumps as it is absorbed or expelled. The scale of such energy lumps (or 'quanta') depends on the wavelength. The larger the wavelength, the greater the sum of energy per photon.
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