Question
Explain the failure of wave-theory to explain the photoelectric effect.
✓
Answer
→The phenomena of interference, diffraction and polarisation have proved that light has a wave nature. According to this, light is a electromagnetic wave consisting of electric and magnetic fields with continuous distribution of energy over the region of space over which the wave is extended.
→According to wave theory the intensity of light is proportional to the square of the amplitude of the light. ( $I \propto A ^2$ )
→Thus, the energy of light is directly related to its intensity i.e. light with higher intensity has more energy, therefore, the maximum energy of the emitted electrons is also higher when the high intensity light falls on the metal surface.
→Thus, according to wave theory the energy of the photoelectron emitted from the metal depends on the intensity of light, which is the opposite from the experimental result.
→According to wave theory if the surface of a metal is exposed to light of sufficient intensity such that the energy received by the electron is greater than the work function of the metal, then electrons will be emitted regardless of the frequency of the light.
→This explanation takes out the presence of a threshold frequency inferred from experimental results.
→According to wave theory light has a continuous distribution of energy as a wave-front so when light is incident on a metal surface the energy absorbed by the electrons is also continuously absorbed by the wave front of the light.
→The number of electrons in a metal is very high, so the energy absorbed by each electron per unit time will be very small.
→Thus, the electron will absorb a very small amount of energy from the light every second and when it becomes equal to the work function, the electron will be emitted. A precise calculation shows that it should take hours for the electrons to be emitted.
→This explanation is also contrary to the phenomenon of spontaneous emission of electrons found by experimental results.
→From this it can be said that the wave theory fails to explain the photoelectric effect.
→According to wave theory the intensity of light is proportional to the square of the amplitude of the light. ( $I \propto A ^2$ )
→Thus, the energy of light is directly related to its intensity i.e. light with higher intensity has more energy, therefore, the maximum energy of the emitted electrons is also higher when the high intensity light falls on the metal surface.
→Thus, according to wave theory the energy of the photoelectron emitted from the metal depends on the intensity of light, which is the opposite from the experimental result.
→According to wave theory if the surface of a metal is exposed to light of sufficient intensity such that the energy received by the electron is greater than the work function of the metal, then electrons will be emitted regardless of the frequency of the light.
→This explanation takes out the presence of a threshold frequency inferred from experimental results.
→According to wave theory light has a continuous distribution of energy as a wave-front so when light is incident on a metal surface the energy absorbed by the electrons is also continuously absorbed by the wave front of the light.
→The number of electrons in a metal is very high, so the energy absorbed by each electron per unit time will be very small.
→Thus, the electron will absorb a very small amount of energy from the light every second and when it becomes equal to the work function, the electron will be emitted. A precise calculation shows that it should take hours for the electrons to be emitted.
→This explanation is also contrary to the phenomenon of spontaneous emission of electrons found by experimental results.
→From this it can be said that the wave theory fails to explain the photoelectric effect.
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