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# The phenomenon of photoelectric emission depends onA) wavelength of incident lightB) work function of surfaceC) nature of surfaceD) all of the above

Last updated date: 20th Jun 2024
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Hint: The photoelectric emission happens when incident light with a certain wavelength excites the free electron on the metal surface. The absorbed energy from the incident photon supplies the excitation energy and the kinetic energy of the emitted electron. This excitation energy is called the work function of the surface.

Formula Used:
The process of the photoelectric emission can be described as
${E_{photon}} = \Phi + {K_{electron}}$
where, ${E_{photon}}$ is the energy of the photon, $\Phi$ is the work function of the metal surface and ${K_{electron}}$ is the kinetic energy of the emitted electron

Step 1:
From eq (1) it is evident that the energy of the photon is the key ingredient for the process of photo emission. The energy of the photon is transferred to the electron to emit from the surface by surpassing the work function and then attaining a kinetic energy of emission.
The energy of photon can be written as
${E_{photon}} = \dfrac{{hc}}{\lambda }$
where, $h$ is the Planck's constant, $c$ is the velocity of light in vacuum and $\lambda$ is the wavelength of the photon.
Thus, the wavelength of the incident light is related to the energy of the photon ${E_{photon}}$ that is an important factor of the photoelectric emission.
So, option A is a correct option.
Step 2:
Again, the work function $\Phi$ also plays an important role in the photoelectric emission. If the absorbed energy from the incident photon, is equal or greater than the work function of the surface then the electron absorbs the energy and gets emitted.
Hence option B is also correct.
Step 3:
The work function of any surface is dependent on the nature of the surface material only.
Hence, option C is also correct.
So, you can get that all the options A, B and C are correct